The radius R of a geosynchronous satellite's orbit is 42.2 x 10⁶ meters, or approximately 6.63 times the radius of the Earth.
The radius R of the orbit of a geosynchronous satellite that circles the Earth can be computed using the principles of circular motion and the law of universal gravitation, considering a geostationary satellite takes 23 hours 56 minutes and 4 seconds to complete one orbit. The gravitational force provides the necessary centripetal force to keep the satellite in orbit. Given the mass of the Earth (me = 5.98 x 10²⁴kg), the mean radius of the earth (R2 = 6.37 x 10⁶ m), and the universal gravitational constant (G = 6.67 x 10⁻¹¹ N m²kg⁻²), the radius R is already provided as 42.2 x 10⁶ m. This is the distance from the center of the Earth to the satellite. When expressed in terms of Earth's radii, this is approximately 6.63 Earth radii, since the radius of the Earth is 6.37 x 10⁶ m.
A horizontal force, F1 = 65 N, and a force, F2 = 12.4 N acting at an angle of θ to the horizontal, are applied to a block of mass m = 3.1 kg. The coefficient of kinetic friction between the block and the surface is μk = 0.2. The block is moving to the right.
Part (a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if θ = 30°.t Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if θ = 30°.
To solve for the normal force, the forces in the vertical direction are balanced, considering gravity and the y-component of force [tex]F_2[/tex] which is approximately 24.211 N. For acceleration, the net force in the horizontal direction is calculated by adding force [tex]F_1[/tex] and the x-component of [tex]F_2[/tex], and subtracting kinetic friction, then divided by the mass which gives [tex]23.14 m/s^2[/tex] approximately.
The problem given requires us to calculate the magnitude of the normal force and the acceleration of the block using Newton's second law and the concepts of kinetic friction.
Part (a)
Firstly, to find the normal force, [tex]F_N[/tex], we need to consider the forces in the vertical (y) direction. The gravitational force (weight) acts downward with a magnitude of m × g, and the vertical component of [tex]F_2[/tex] acts upward. Additionally, the normal force acts upward, balancing these forces.
Thus, we have:
[tex]F_N + F_{2y} - m \times g = 0[/tex]
Where:
[tex]F_N[/tex] is the normal force
[tex]F_{2y} = F_2[/tex] × sin(θ) (the vertical component of [tex]F_2[/tex])
m × g is the weight of the block, m is the mass and g is the acceleration due to gravity, which is approximately [tex]9.81 m/s^2[/tex]
Plugging in the known values:
[tex]F_N[/tex] + 12.4 × sin(30°) - 3.1 × 9.81 = 0
[tex]F_N[/tex] = 3.1 × 9.81 - 12.4 × 0.5
[tex]F_N[/tex] = 30.411 - 6.2
[tex]F_N[/tex] = 24.211 N (approximately)
Part (b)
For the acceleration, a, we consider the forces in the horizontal (x) direction. The net force is the sum of the horizontal component of [tex]F_2[/tex] and [tex]F_1[/tex] minus the kinetic friction force, [tex]F_k[/tex].
The kinetic friction force, [tex]F_k[/tex], is given by:
[tex]F_k = \mu_k \times F_N[/tex]
And hence:
[tex]F_{net}[/tex] = [tex]F_1 + F_{2x} - F_k[/tex]
Using [tex]F_N[/tex] from the first section to find [tex]F_k[/tex] and then [tex]F_{net}[/tex], we get:
[tex]F_{net}[/tex] = 65 + 12.4 × cos(30°) - 0.2 × 24.211
Using Newton's second law, a = [tex]F_{net}[/tex] / m:
a = (65 + 12.4 × 0.866 - 0.2 × 24.211) / 3.1
a = (65 + 10.7424 - 4.8422) / 3.1
a = 71.7424 / 3.1
a = [tex]23.14 m/s^2[/tex] (approximately)
The potential difference V(t) between the parallel plates of a capacitor is instantaneously increasing at a rate of 107 V/s. What is the displacement current (in mA) between the plates if the separation of the plates is 1.06 cm and they have an area of 0.174 m2?
To solve this problem it is necessary to apply the related concepts to the scalar value of displacement current, which can be expressed in terms of electric flux as
[tex]I_d = \epsilon_0 \frac{d\Phi_E}{dt}[/tex]
Where,
[tex]\epsilon_0[/tex] = Permitibitty of free space constant
[tex]\Phi_E[/tex] = Magnetic flux
t = time
We know as well that the Flux can be expressed as
[tex]\Phi = EA[/tex]
Here
A= Cross-sectional area
E = Electric Potential in a Uniform electric field
At the same time the electric potential is expressed in terms of Voltage and distance, that is
[tex]E = \frac{V}{d}[/tex]
Using this equation we have then that
[tex]I_d = \epsilon_0 \frac{d\Phi_E}{dt}[/tex]
[tex]I_d = \epsilon_0 \frac{d(EA)}{dt}[/tex]
[tex]I_d = \epsilon_0*A (\frac{d(E)}{dt})[/tex]
[tex]I_d = \epsilon_0*A (\frac{d(V)}{dt*d})[/tex]
[tex]I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})[/tex]
According to our values we have that
[tex]\frac{dV}{dt} = 107V/s[/tex]
[tex]A = 0.174m^2[/tex]
[tex]d = 1.06*10^{-3}m[/tex]
[tex]\epsilon = 8.85418^{-12} m^{-3}kg^{-1}s^4A^2[/tex]
Replacing,
[tex]I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})[/tex]
[tex]I_d = \frac{(8.85418^{-12})*(0.174)}{1.06*10^{-3}} (107)[/tex]
[tex]I_d = 7.565*10^{-8}A[/tex]
Therefore the displacement current is [tex]7.565*10^{-5}mA[/tex]
The displacement current (in mA) between these plates is equal to [tex]1.56 \times 10^{-5}\;mA[/tex]
Given the following data:
Rate of increment = 107 V/s.Distance = 1.06 cm to m = [tex]1.06 \times 10^{-2}\;m[/tex]Area = 0.174 [tex]m^2[/tex].Scientific data:
Permittivity of free space = [tex]8.854 \times 10^{-12}[/tex]How to calculate the displacement current.Mathematically, the displacement current (in mA) between the plates in an electric field is given by this formula:
[tex]I_d=\frac{\epsilon _o A}{d} (\frac{d(v)}{dt} )\\\\[/tex]
Where:
[tex]\epsilon_o[/tex] is the permittivity of free space.d is the distance.A is the area.[tex]\frac{d(v)}{dt}[/tex] is the rate of change.Substituting the given parameters into the formula, we have;
[tex]I_d=\frac{8.854 \times 10^{-12} \times 0.174 \times 107}{1.06 \times 10^{-2}} \\\\I_d=1.56 \times 10^{-8}\\\\I_d=1.56 \times 10^{-5}\;mA[/tex]
Read more on potential difference here: brainly.com/question/4313738
Arrange the following kinds of electromagnetic radiation in order of increasing wavelength: infrared, green light, red light, radio waves, X rays, ultraviolet light. Rank from shortest to longest. To rank items as equivalent, overlap them.
Answer:
X rays, ultraviolet light, green light, red light, infrared radio waves
Explanation:
The electromagnetic spectrum can be thought of as being arranged either in decreasing frequency or increasing wavelength. The full spectrum is as follows, from the shortest wavelength to the longest:
Gamma rays, X rays, ultraviolet light, violet light, indigo light, blue light, green light, yellow light, orange light, red light, infrared, radio waves, radar waves, microwaves, television waves, radio waves.
The electromagnetic radiation types can be arranged from shortest to longest wavelength as follows: X rays, ultraviolet light, green light, red light, infrared, and radio waves.
Explanation:The electromagnetic radiation types provided can be arranged in order of increasing wavelength as follows: X rays, ultraviolet light, green light, red light, infrared, radio waves. This order represents the electromagnetic spectrum from shortest to longest wavelengths. In simpler terms, X rays have the shortest wavelength, followed by ultraviolet light, green light, red light, infrared, and finally radio waves with the longest wavelength. Wavelength and frequency are inversely related - the shorter the wavelength, the higher the frequency.
Learn more about Electromagnetic Radiation here:https://brainly.com/question/2334706
#SPJ6
A liquid of density 1230 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.39 m/s and the pipe diameter d1 is 11.5 cm. At Location 2, the pipe diameter d2 is 15.3 cm. At Location 1, the pipe is Δy=9.59 m higher than it is at Location 2. Ignoring viscosity, calculate the difference ΔP between the fluid pressure at Location 2 and the fluid pressure at Location 1.
To develop this problem it is necessary to use the continuity equations and Bernoullie's theorem.
It is known from Bernoullie's theorem that
[tex]P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2[/tex]
Where
P = Pressure
g = Gravity
h= Height
v = Velocity
[tex]\rho[/tex] = Density
On the other hand we have that the continuity equation is given by
[tex]A_1v_1 = A_2 v_2[/tex]
Where A is the Cross-sectional area and v the velocity.
For our values we know that
[tex]A_1v_1 = A_2 v_2[/tex]
[tex](\frac{\pi d_1^2}{4})v_1 =(\frac{\pi d_2^2}{4})v_2[/tex]
[tex]d_1^2v_1=d_2^2v_2[/tex]
[tex](11.5cm)^2(9.39)=(15.3)^2v_2[/tex]
[tex]v_2 = 5.305m/s[/tex]
Using Bernoulli's expression we can now find the pressure difference,
[tex]P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2[/tex]
[tex]P_1-P_2=-\rho gh_1-\frac{1}{2}\rho v_2^2 +\rho gh_2+\frac{1}{2} \rho v_2^2[/tex]
[tex]P_1-P_2 = \rho g (h_1-h_2)+\frac{1}{2}\rho(v_1^2-v_2^2)[/tex]
[tex]P_1-P_2 = (1.3*10^3)(9.8)(9.59)+\frac{1}{2}(1.3*10^3)((9.39)^2-(5.305)^2)[/tex]
[tex]P_1-P_2 = 1.612*10^5Pa[/tex]
illustrates an Atwood's machine. Let the masses of blocks A and B be 6.00 kg and 3.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.220 kg⋅m2, and the radius of the wheel be 0.120 m. There is no slipping between the cord and the surface of the wheel.
Answer:
The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.
Explanation:
Given that,
Mass of block A = 6.00 kg
Mass of block B = 3.00 kg
Moment of inertia = 0.220 kg.m²
Radius = 0.120 m
Suppose we need to find the the magnitude of the linear acceleration of block A
Let a is the acceleration of the blocks.
Let [tex]T_{a}[/tex] and [tex]T_{b}[/tex] are the tension in the A and B cord.
According to figure,
We need to calculate the magnitude of the linear acceleration of block A
Net force acting on block A,
[tex]F_{A}=m_{A}g-T_{A}[/tex]
[tex]m_{A} a=m_{A}g-T_{A}[/tex]
[tex]T_{A}=m_{A}g-m_{A}a[/tex]...(I)
Net force acting on block B,
[tex]F_{B}=T_{B}-m_{B}g[/tex]
[tex]m_{B}a=T_{B}-m_{B}g[/tex]
[tex]T_{B}=m_{B}a+m_{B}g[/tex]...(II)
Net torque acting on pulley
[tex]T_{net}=I\times\alpha[/tex]
[tex]T_{A}r-T_{B}r=I\times \dfrac{a}{r}[/tex]
[tex]T_{A}-T_{B}=I\times\dfrac{a}{r^2}[/tex]
[tex]m_{A}g-m_{A}a-(m_{B}g+m_{B}a)=I\times\dfrac{a}{r^2}[/tex]
[tex]g(m_{A}-m_{B})-a(m_{A}+m_{B})=I\times\dfrac{a}{r^2}[/tex]
[tex]g(m_{A}-m_{B})=I\times\dfrac{a}{r^2}+a(m_{A}+m_{B})[/tex]
[tex]g(m_{A}-m_{B})=a(\dfrac{I}{r^2}+(m_{A}+m_{B}))[/tex]
[tex]a=\dfrac{g(m_{A}-m_{B})}{(\dfrac{I}{r^2}+(m_{A}+m_{B}))}[/tex]
Put the value into the formula
[tex]a=\dfrac{9.8\times(6.00-3.00)}{\dfrac{0.220}{(0.120)^2}+(6.00+3.00)}[/tex]
[tex]a=1.21\ m/s^2[/tex]
Hence, The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.
A produce distributor uses 800 packing crates a month, which it purchases at a cost of $10 each. The manager has assigned an annual carrying cost of 35 percent of the purchase price per crate. Ordering costs are $28. Currently the manager orders once a month.
Answer:
$364.29
Explanation:
given,
Packing of crates per month (u)= 800
annual carrying cost of 35 percent of the purchase price per crate.
Ordering cost(S) = $ 28
D = 800 x 12 = 9600 crates/year
H = 0.35 P
H = 0.35 x $10
H = $3.50/crate per yr.
Present Total cost
= [tex]\dfrac{800}{2}\times 3.50 + \dfrac{9600}{800}\times 28[/tex]
= 1400 + 336
= $ 1,736
[tex]Q_0 = \sqrt{\dfrac{2DS}{H}}[/tex]
[tex]Q_0 = \sqrt{\dfrac{2\times 9600 \times 28}{3.50}}[/tex]
[tex]Q_0 =\$ 391.92[/tex]
Total cost at EOQ
= [tex]\dfrac{391.92}{2}\times 3.50 + \dfrac{9600}{391.92}\times 28[/tex]
= 685.86 + 685.85
= $ 1,371.71
the firm save annually in ordering and carrying costs by using the EOQ
= $ 1,736 - $ 1,371.71
= $364.29
The question focuses on Economic Order Quantity (EOQ) in inventory management for a produce distributor's packing crates. EOQ helps determine the optimal number of units a company should add to its inventory to minimize total holding and setup costs.
Explanation:The subject of this question is
Economic Order Quantity (EOQ)
, which is an operational efficiency measure used in inventory management. It refers to the number of units that a company should add to its inventory at a time to minimize total holding and setup costs.
In this case, the produce distributor uses 800 crates a month, purchased at $10 each. The annual carrying cost is 35% of the purchase price per crate, and the ordering cost is $28. Currently, the distributor orders once a month. Hence, to determine the optimal number of orders, or the EOQ, various factors such as ordering cost, holding cost, and demand rate need to be considered.
This cost minimization issue can be mathematically represented and solved using the EOQ formula: EOQ = √((2DS)/H) where D is the demand rate, S is the setup (or order) cost, and H is the holding (or carrying) cost.
Learn more about EOQ here:https://brainly.com/question/34285831
#SPJ3
When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature (37° C) by warming up the water in your stomach. Could drinking ice water, then, substitute for exercise as a way to "burn calories?" Suppose you expend 286 kilocalories during a brisk hour-long walk. How many liters of ice water (0° C) would you have to drink in order to use up 286 kilocalories of metabolic energy? For comparison, the stomach can hold about 1 liter.
Answer:
7.72 Liters
Explanation:
normal body temperature = T_body =37° C
temperature of ice water = T_ice =0°c
specfic heat of water = c_{water} =4186J/kg.°C
if the person drink 1 liter of cold water mass of water is = m = 1.0kg
heat lost by body is Qwater =mc_{water} ΔT
= mc{water} ( T_ice - T_body)
= 1.0×4186× (0 -37)
= -154.882 ×10^3 J
here negative sign indicates the energy lost by body in metabolic process energy expended due to brisk - hour long walk is Q_{walk} = 286 kilocalories
= 286×4186J
so number of liters of ice water have to drink is
n×Q_{water} =Q_{walk} n= Q_{walk}/ Q_{water}
= 286×4186J/154.882×10^3 J
= 7.72 Liters
Answer:
[tex]V=7.73\ L[/tex]
Explanation:
Given:
Initial temperature of water, [tex]T_i=0^{\circ}C[/tex]
final temperature of water, [tex]T_f=37^{\circ}C[/tex]
energy spent in one hour of walk, [tex]286\ kilocal=(286\times 4186)\ J[/tex]
volumetric capacity of stomach, [tex]V=1\ L[/tex]
Now, let m be the mass of water at zero degree Celsius to be drank to spend 286 kilo-calories of energy.
[tex]\therefore Q=m.c_w.\Delta T[/tex] .....................................(1)
where:
m = mass of water
Q = heat energy
[tex]c_w=4186\ J\ (specific\ heat\ of\ water)[/tex]
[tex]\Delta T[/tex]= temperature difference
Putting values in the eq. (1):
[tex]286\times 4186=m\times 4186\times 37[/tex]
[tex]m=7.73\ kg[/tex]
Since water has a density of 1 kilogram per liter, therefore the volume of water will be:
[tex]V=7.73\ L[/tex]
Two in-phase loudspeakers are placed along a wall and are separated by a distance of 4.00 m. They emit sound with a frequency of 514 Hz. A person is standing away from the wall, in front of one of the loudspeakers. What is the closest distance from the wall the person can stand and hear constructive interference? The speed of sound in air is 343 m/s.
Multiple choice:
1.64 m
1.15 m
0.344 m
0.729 m
The smallest distance from the wall where a listener could hear constructive interference from two in-phase loudspeakers emitting sound at a frequency of 514 Hz would be 0.668 m, under the conditions given in the problem.
Explanation:The phenomenon being described here is constructive interference, which occurs when two waves overlap so that their combined effect is stronger than their individual effects. In this scenario, we calculate the path difference, which is the difference in the path traveled by the sound from the two loudspeakers. Constructive interference occurs when the path difference corresponds to an integer number of wavelengths.
First, let's calculate the wavelength of the sound: λ = speed of sound / frequency. So, λ = 343 m/s / 514 Hz = 0.668 m.
For constructive interference, the path difference should be Nλ, where N is an integer. Since we're looking for the smallest possible distance, we should take N = 1. So, the distance from the wall where the listener stands (d = path difference) should be 1 × λ = 1 × 0.668 m = 0.668 m. But none of the options given match this value. Therefore, none of the options given are correct, if we correctly followed the conditions given in the problem.
Learn more about Constructive Interference here:https://brainly.com/question/31857527
#SPJ12
The person standing in front of the loudspeakers, which are emitting sound waves with a frequency of 514 Hz, will experience constructive interference at a distance closest to 1.15 m from one of the speakers.
Explanation:In this case, the person standing in front of the loudspeakers will experience constructive interference - where the waves from the two loudspeakers are in-phase and add up to make a louder sound - at distances that are multiples of one wavelength away from one speaker. The wavelength (λ) of the sound can be calculated using the formula λ = v / f, where v is the speed of sound and f is the frequency. Substituting the given values, λ = 343 m/s / 514 Hz = 0.667 m.
The student will hear constructive interference at distances of nλ/2 from the speaker, where n is an integer. The smallest non-zero value for n is 1, so the closest such point will be 0.667 m / 2 = 0.334 m. However, this is less than the smallest distance in the multiple choice answers, which suggest the answer should be in metres. The next value for n is 2, which gives a distance of 0.667 m, still less than the smallest multiple choice answer. The next value for n is 3, which gives a distance of 1.001 m, which falls between the 1st and 2nd options so the answer should correspond to the closest choice which is 1.15 m.
Learn more about Constructive Interference here:https://brainly.com/question/31857527
#SPJ12
A tub containing 53 kg of water is placed in a farmer's canning cellar, initially at 10∘C. On a cold evening the cellar loses thermal energy through the walls at a rate of 1200 J/s. Without the tub of water, the fruit would freeze in 4 h (the fruit freezes at −1∘C because the sugar in the fruit lowers the freezing temperature). Part A By what time interval does the presence of the water delay the freezing of the fruit? The specific heat of water is c = 4180 J/kg⋅∘C.
To solve this problem it is necessary to apply the concepts related to heat transfer and power depending on energy and time.
By definition we know that the heat loss of water is given by
[tex]Q = mc_w*\Delta T+m*L_f[/tex]
Where,
m = mass
[tex]c_w =[/tex] Specific Heat of Water
T = Temperature
[tex]L_f =[/tex]Latent heat of fusion [tex] \rightarrow[/tex] Heat of fusion for water at 0°C is [tex]3.35*10^5J/Kg[/tex]
Our values are given as,
m=53 kg
[tex]C=4180 J/kg\°C[/tex]
[tex]\Delta T=10-(-1)=11[/tex]
Replacing we have,
[tex]Q=53*4180*6+53*3.35*10^5[/tex]
[tex]Q = 19084240J[/tex]
Power can be defined as
[tex]P = \frac{Q}{t}[/tex]
Re-arrange to find t,
[tex]t = \frac{Q}{P}[/tex]
[tex]t = \frac{19084240}{1200}[/tex]
[tex]t = 15903.53s \approx 265 min \aprox 4.41h[/tex]
Therefore the time interval is 4.41h
The question revolves around the principle of thermal energy transfer and the specific heat of water. The water’s high specific heat absorbs a large amount of heat from the cellar and hence delays the freezing of the fruit by an additional 0.57 hours.
Explanation:The subject of this question involves the principles of thermal Energy and heat transfer, specifically regarding the specific heat of water. The water in the tub has the ability to retain heat and slow down the cooling process of the space due to water's high specific heat capacity which is 4180 J/kg⋅°C. This means that water can absorb a lot of heat before its temperature rises.
Firstly, we need to calculate how much heat is the water in the tub able to absorb before its temperature reaches -1 °C (the freezing point of the fruit). This can be calculated with the formula for heat transfer Q = mcΔT, where m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. In this case, ΔT is 10 - (-1) = 11 °C, so Q = 53 kg * 4180 J/kg⋅°C * 11 °C = 2,444,840 J of heat.
The cellar loses heat at a rate of 1200 J/s, which means to lose 2,444,840 J of heat it would require 2,444,840 J / 1200 J/s = 2,037.3 seconds, which is approximately 0.57 hours. So, the presence of the water in the tub delays the freezing of the fruit by an additional 0.57 hours.
Learn more about Thermal Energy & Heat Transfer here:https://brainly.com/question/16961806
#SPJ11
A solid uniform sphere and a thin-walled, hollow sphere have the same mass M and radius R. If they roll without slipping up a ramp that is inclined at an angle β above the horizontal and if both have the same vcm before they start up the incline, which object reaches the greater height, or do both of them reach the same height?
Answer:Hollow sphere
Explanation:
Given
same mass for solid and hollow sphere
same [tex]v_{cm}[/tex] before they start up incline
Moment of inertia of solid Sphere
[tex]I_1=\frac{2}{5}Mr^2[/tex]
Moment of inertia of hollow sphere
[tex]I_2=\frac{2}{3}Mr^2[/tex]
Conserving Energy at bottom and top point for solid sphere
kinetic energy +Rotational Energy=Potential energy
[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\omega ^2=mgh_1[/tex]
for pure rolling [tex]v_{cm}=\omega r[/tex]
[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{5}Mr^2=Mgh_1[/tex]
[tex]\frac{7}{10}Mv_{cm}^2=Mgh_1[/tex]
[tex]h_1=\frac{7v_{cm}^2}{10g}[/tex]
For hollow sphere
[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{3}Mr^2=Mgh_2[/tex]
[tex]h_2=\frac{5v_{cm}^2}{6g}[/tex]
therefore height gained by hollow sphere is more
The hollow sphere reached the greatest height.
Height reached by each object
The height reached by each object is determined by applying the principle of conservation of energy as show below;
K.E = P.E
¹/₂mv² + ¹/₂Iω² = mgh
where;
I moment of inertia ω is angular speed = v/rHeight reached by solid sphereI = ²/₅mr²
¹/₂mv² + ¹/₂Iω² = mgh
¹/₂mv² + ¹/₂(²/₅mr²)(v/r)² = mgh
¹/₂v² + ¹/₅v² = gh
7v² = 10gh
h = 7v²/10g
h = 0.7(v²/g)
Height reached by hollow sphereI = ²/₃mr²
¹/₂mv² + ¹/₂Iω² = mgh
¹/₂mv² + ¹/₂(²/₃mr²)(v/r)² = mgh
¹/₂v² + ¹/₃v² = gh
5v² = 6gh
h = 5v²/6g
h = 0.83(v²/g)
Thus, the hollow sphere reached the greatest height.
Learn more about conservation of energy here: https://brainly.com/question/166559
A typical tornado can be envisioned as a cylinder of height 640 m, with a diameter of about 230 m (note: the visible funnel cloud is usually somewhat less than the actual diameter). The rotation in the tornado is approximately a `solid body rotation', meaning that it rotates very much like a solid cylinder would, even though air is a gas. If the outer edge of the cylinder has a speed of 51 m/s, and the air has a density of 0.97 kg/m3, determine the kinetic energy contained by the tornado.
Answer:
16771720740.20324 J
Explanation:
[tex]\rh0[/tex] = Density = 0.97 kg/m³
V = Volume = [tex]\pi r^2h[/tex]
d = Diameter of cylinder = 230 m
r = Radius = [tex]\frac{d}{2}=\frac{230}{2}=115\ m[/tex]
h = Height of the cylinder = 640 m
v = Velocity of cylinder = 51 m/s
Mass of object is given by
[tex]m=\rho V\\\Rightarrow m=0.97\times \pi 115^2\times 640\\\Rightarrow m=25792727.013\ kg[/tex]
Moment of inertia of a cylinder
[tex]I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}\times 25792727.013\times 115^2\\\Rightarrow I=170554407373.4625\ kgm^2[/tex]
Angular speed
[tex]\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{51}{115}\ rad/s[/tex]
Kinetic energy is given by
[tex]K=\frac{1}{2}I\omega^2\\\Rightarrow K=\frac{1}{2}\times 170554407373.4625\times \left(\frac{51}{115}\right)^2\\\Rightarrow K=16771720740.20324\ J[/tex]
The kinetic energy contained by the tornado is 16771720740.20324 J
A rescue team is searching for Andrew, a geologist who was stranded while conducting research in the mountains of Colorado. The team uses electronic listening devices in order to detect any shouts for help. The sound waves from his shouts reaching the base camp can be approximated by a sinusoidal wave with a frequency f = 530 Hz and displacement amplitude A = 2.00×10−8 m , where the sound wave properties are valid at the base camp where the measurements are being made. What sound intensity level will the rescue team measure from the frightened researcher? Assume the speed of sound is v=328m/s and the density of air rho = 1.34 kg/m3
Final answer:
To calculate the sound intensity level, we use the formula I = (A² * ρ * v * f)/2. Plugging in the given values, the sound intensity level will be approximately 4.32 × 10⁻⁶ W/m².
Explanation:
To calculate the sound intensity level, we can use the formula:
I = (A² * ρ * v * f)/2
Where:
I is the sound intensity level,
A is the displacement amplitude,
ρ is the density of air,
v is the speed of sound,
f is the frequency of the sound wave.
Plugging in the given values, we have:
I = (2.00×10⁻⁸ m)² * 1.34 kg/m³ * 328 m/s * 530 Hz)/2
Simplifying the expression, the sound intensity level will be approximately 4.32 × 10⁻⁶ W/m².
A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.
(a) What impulse acts on the ball during the contact?
(b) If the ball is in contact with the floor for 0.020 s, what is the magnitude of the average force on the floor from by the ball? Neglect air resistance
Explanation:
Given that,
Mass of the ball, m = 1.2 kg
Initial speed of the ball, u = 10 m/s
Height of the floor from ground, h = 32 m
(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8\times 32}[/tex]
v = -25.04 m/s (negative as it rebounds)
The impulse acting on the ball is equal to the change in momentum. It can be calculated as :
[tex]J=m(v-u)[/tex]
[tex]J=1.2\times (-25.04-10)[/tex]
J = -42.048 kg-m/s
(b) Time of contact, t = 0.02 s
Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :
[tex]J=\dfrac{F}{t}[/tex]
[tex]F=J\times t[/tex]
[tex]F=42.048\times 0.02[/tex]
F = 0.8409 N
Hence, this is the required solution.
Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck.Which box ends up moving faster?a. Box 1b. Box 2c. Same
Answer:
a. Box 1
Explanation:
Hi there!
The momentum of the system box-ball is conserved in both cases because there is no external force applied on the system.
The momentum of the system is calculated as the sum of the momenta of each object that composes the system. The momentum is calculated as follows:
p = m · v
Where:
p = momentum.
m = mass.
v = velocity.
Then, the momentum of the system before and after the collision will be:
System ball - box 1
initial momentum = final momentum
mb · vb + m1 · v1 = mb · vb´ + m1 · v1´
Where:
mb = mass of the ball.
vb = veloctiy of the ball.
m1 = mass of box 1.
v1 = velocity of box 1.
vb´ = final velocity of the ball.
v1´ = final velocity of box 1.
Since the initial velocity of the box is zero:
mb · vb = mb · vb´ + m1 · v1´
Solving for v1´
mb · vb - mb · vb´ = m1 · v1´
mb · (vb - vb´) = m1 · v1´
mb · (vb - vb´) / m1 = v1´
Since vb´ is negative because the ball bounces back, then:
mb · (vb + vb´) / m1 = v1´
Now let´s express the momentum of the system ball - box 2
System ball -box 2
mb · vb + m2 · v2 = (mb + m2) · v2´
Since v2 = 0
mb · vb = (mb + m2) · v2´
Solving for v2´:
mb · vb / (mb + m2) = v2´
Comparing the two expressions:
v2´ = mb · vb / (mb + m)
v1´ = mb · (vb + vb´) / m
In v1´ the numerator is greater than the numerator in v2´ because
vb + vb´> vb
In v2´ the denominator is greater than the denominator in v1´ because
mb + m > m
then v1´ > v2´
Box 1 ends up moving faster than box 2
Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and kinetic and potential energy effects are negligible. Assuming the ideal gas model for the air, determine the maximum theoretical pressure at the exit, in lbf/in.2 p2,max = lbf/in.2
To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.
By definition the relationship between pressure and temperature is given by
[tex](\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}[/tex]
Here
P = Pressure
T = Temperature
[tex]\gamma =[/tex]The ratio of specific heats. For air normally is 1.4.
Our values are given as,
[tex]P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K[/tex]
Therefore replacing we have,
[tex](\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}[/tex]
[tex](\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}[/tex]
Solving for [tex]P_2,[/tex]
[tex]P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}[/tex]
[tex]P_2 = 44.15Lbf/in^2[/tex]
Therefore the maximum theoretical pressure at the exit is [tex]44.15Lbf/in^2[/tex]
To determine the maximum theoretical pressure at the exit of the compressor, we can use the ideal gas law. Given the initial pressure and temperature, as well as the final temperature, we can calculate the maximum theoretical pressure using the equation p2,max = (p1 * T2) / T1.
Explanation:To determine the maximum theoretical pressure at the exit of the compressor, we can use the ideal gas law. The ideal gas law is given by the equation pV = nRT, where p is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We can rearrange this equation to solve for the pressure:
p = (nRT) / V
Since the volume and the number of moles are constant, we can rewrite the equation as:
p1 / p2 = T1 / T2
where p1 and T1 are the initial pressure and temperature, and p2 and T2 are the final pressure and temperature.
Substituting the given values, we have:
p2,max = (p1 * T2) / T1
Plugging in the values from the problem statement, we have:
p2,max = (15 lbf/in.2 * 275°F) / 80°F
p2,max = 51.56 lbf/in.2
Learn more about Ideal gas law here:https://brainly.com/question/30458409
#SPJ11
A vertical spring has a length of 0.25 m when a 0.175 kg mass hangs from it, and a length of 0.775 m when a 2.075 kg mass hangs from it. gA. What is the force constant of the spring, in newtons per meter?
B. What is the unloaded length of the spring, in centimeters?
Answer:
A) 35.5N/m b) 20.1cm
Explanation:
Using Hooke's law;
F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters
For the first body, m*g = K * (0.25- li)
Where li is the initial length of the spring
0.175*9.81 = k(0.25-li)
1.72 = k(0.25-li) as equation 1
For the second body, m *g = K* ( 0.775-li)
2.075*9.81 = k (0.775-li) equation 2
20.36 = k(0.775-li)
Make li subject of the formula;
li = 0.775 - 20.36/k
Substitute for li in equation 1
1.72 = k(0.25- (0.775 - 20.36/k))
1.72 = k ( 0.25 - 0.775 + 20.36/k)
Open the bracket with k
1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)
Collect the like terms:
1.72 - 20.36 = - 0.525k
- 18.64 = -0.525k
Divide both side by -0.525
-18.64/-0.525 = -0.525/-0.525k
K = 35.5N/m
B) substitute for k in using
li = 0.775 - 20.36/k
li = 0.775 - 20.36/35.5
li = 0.775 - 0.574
li = 0.201 in meters
li = 0.201 * 100 centimeters = 20.1cm
(A) The force constant of the spring is obtained as 34.66 N/m
(B) The unloaded length of the spring is 20 cm.
Hooke's Law(A) According to Hooke's law;
[tex]F = kx[/tex]
Where 'F' is the force needed to extend or compress the spring, 'k' is the force constant and 'x' is the extended length or compressed length.
Here, the gravitational force of the mass provides the necessary force for the spring to elongate.
Therefore, we can say that in the case of first mass;
[tex]k\,(0.25-l_0) = m_1g = (0.175\,kg \times 9.8\,m/s^2)[/tex]
Where [tex]l_0[/tex] is the unloaded length of the spring.
[tex]\implies k\,(0.25-l_0)=1.715\,N[/tex]
In the case of the second mass, we can write;
[tex]k\,(0.775\,m-l_0) = m_2g = (2.075\,kg \times 9.8\,m/s^2)[/tex]
[tex]\implies k\,(0.775\,m-l_0) = 20.335\,N[/tex]
From both these equations, we can write;
[tex]k\,(0.775\,m-l_0-0.25\,m+l_0)=20.335\,N- 1.715\,N[/tex]
[tex]\implies0.525\, k=18.2\,N[/tex]
[tex]\therefore k=\frac{18.2\,N}{0.525\,m}=34.66\,N/m[/tex]
(B) Applying the value of 'k' in any of the equations for force of given masses, we get;
[tex]k\,(0.25-l_0)=1.715\,N[/tex]
[tex]\implies (34.66\,N/m)\times\,(0.25-l_0)=1.715\,N[/tex]
[tex]\implies 8.665N-34.66\,l_0=1.715\,N[/tex]
[tex]\implies \,l_0=\frac{-6.95N}{-34.66} =0.2\,m=20\,cm[/tex]
Learn more about Hooke's law here:
https://brainly.com/question/12253978
A baseball is pitched horizontally toward home plate with a velocity of 110 km/h. In which of the following scenarios does the baseball have the largest change in momentum?
a. The baseball is hit straight back to the pitcher at a speed of 110 km/h.
b. The catcher catches the ball.
c. The ball is popped straight up at a speed of 110 km/h.
d. The catcher catches the ball, and the ball is popped straight up at a speed of 110 km/h have the same change in momentum.
Answer:
option A
Explanation:
given,
velocity to hit home plate = 110 Km/h
a) When the baseball is hit straight back
Assuming the momentum before hitting be P and after hitting the bal is also equal to P.
change in momentum = P - (-P) = 2P
b) When the catcher catches the ball the change in momentum is equal to zero.
c) when the ball is popped up the change in momentum
p_x = P and P_y = P
resultant momentum = [tex]\sqrt{P^2+ P^2} = \sqrt{2}\ P[/tex]
the maximum change in momentum will be in case A
The correct answer is option A
Final answer:
The scenario in which the baseball has the largest change in momentum is when it is hit straight back to the pitcher at a speed of 110 km/h.
Explanation:
The scenario in which the baseball has the largest change in momentum is when it is hit straight back to the pitcher at a speed of 110 km/h. This is because when the baseball is hit straight back, it completely reverses its direction of motion, resulting in a large change in momentum. The other scenarios, such as the baseball being caught by the catcher or popped straight up, do not involve a complete reversal of direction, so the change in momentum is smaller.
A long steel I-beam used in bridge construction has a length of 15.0 m. The beam may be exposed to temperatures as low as −23°C in the winter and as high as 32°C in the summer. What is the difference in length of the beam between these two temperature extremes? (Give the absolute value of the difference. Enter your answer in mm.)
To solve this problem it is necessary to apply the concepts of thermal expansion. Thermal expansion can be expressed in longitudinal terms such as
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
Where,
[tex]\alpha =[/tex] Thermal Expanssion Coefficient
[tex]L_0 =[/tex] Initial Length
[tex]\Delta T =[/tex] Change in Temperature
Our values are given as
[tex]\alpha = 11*10^{-6}/\°C \rightarrow[/tex] from Steel
[tex]L_0 = 15m[/tex]
[tex]T_1 = -23\°C[/tex]
[tex]T_2 = 32\°C[/tex]
Replacing we have that,
[tex]\Delta L = L_0 \alpha \Delta T[/tex]
[tex]\Delta L = (17)(11*10^{-6})(32-(-21))[/tex]
[tex]\Delta L = 0.009911m[/tex]
[tex]\Delta L = 9.911mm[/tex]
Therefore the difference in length of the beam between these two temperature extremes is 9.911mm
A steel ball, with diameter = 3 mm and density = 7600 kg/m3 is dropped in a liquid with density = 1200 kg/m3 . The average time for the ball to travel a distance of 0.50 m is 10.0 s. a. Does the ball rise or fall in the fluid? Explain b. Determine the fluid viscosity.
Answer
given,
diameter of the steel ball = 3 m = 0.003 m
density of the steel = 7600 kg/m³
density of liquid = 1200 kg/m³
distance travel by the ball = 0.5 m
time = t = 10 s
average velocity =[tex]v = \dfrac{0.5}{10}[/tex]
v = 0.05 m/s
a) density of water is less than ball so, ball will fall in the fluid.
gravitational force is equal to buoyancy force and the drag force
[tex]F_g = F_b + F_d[/tex]
[tex]F_g = \rho_g Vg[/tex]
Density of ball = ρ_s
V is the volume ball
buoyancy force
[tex]F_b = \rho_f Vg[/tex]
[tex]F_b = \rho_f V g[/tex]
drag force
[tex]F_d =3 \pi \mu d v[/tex]
[tex]F_g = F_b + F_d[/tex]
[tex] \rho_g Vg = \rho_f V g+ 3 \pi \mu d v[/tex]
[tex] (\rho_g - \rho_f)Vg = 3 \pi \mu d v[/tex]
[tex]V= \dfrac{1}{6}\pi d^3[/tex]
[tex] (\rho_g - \rho_f). \dfrac{1}{6}\pi d^3.g = 3 \pi \mu d v[/tex]
[tex]\mu = \dfrac{(\rho_s-\rho_f)d^2g}{18 v}[/tex]
[tex]\mu = \dfrac{(7600 -1200)\times 0.003^2\times 9.8 }{18 \times 0.05}[/tex]
μ = 0.63 kg m/s
Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates.
These are called geosynchronous orbits. The altitude of a geosynchronous orbit is 3.58×107m(≈22,000miles).
Part A
What is the period of a satellite in a geosynchronous orbit?
Part B
Find the value of g at this altitude.
Part C
What is the weight of a 2000 kg satellite in a geosynchronous orbit?
Answer:
a) 24 Hs. b) 0.224 m/s² c) 448 N
Explanation:
a) As satellites in a geosynchronous orbits, stay directly over a point fixed on the Equator while the Earth rotates, the only way that this can be possible, if the period of the satellite (time to complete a full orbit) is equal to the time that the Earth uses to complete a spin itself, which is exactly one day.
b)
The value of g, is just the acceleration due to the gravitational attraction between the satellite and the Earth.
According the Universal Law of Gravitation, this force can be written in this way:
Fg = ms . a = G me. ms / (re+rs)² ⇒a=g= G me / (re + rs)²
Replacing by the values of G, me, re, and rs, we get:
g = 6.67. 10⁻¹¹ . 5.97.10²⁴ / (6.37 10⁶ + 3.58.10⁷)² m/s²
g= 0.224 m/s²
c) If we call "weight" to the magnitude of the gravitational force on the satellite (as we do with masses on Earth), we can find this value, just solving the equation for Fg, as follows:
Fg = G me . ms / (re + rs)²
Replacing by the values, we find:
Fg = 448 N
A satellite in geosynchronous orbit has a period of 24 hours. The gravitational acceleration at this altitude is roughly 0.224 m/s^2. The weight of a 2000 kg satellite in a geosynchronous orbit is roughly 448 N.
Explanation:A satellite in a geosynchronous orbit has a period of 24 hours, which equates to how long it takes the Earth to rotate once on its axis. The value of gravitational acceleration g at an altitude of a geosynchronous orbit can be calculated using the formula g = G × (M / (R+h)^2), where G is the gravitational constant, M is the mass of the Earth, R is the radius of the Earth, and h is the altitude of the satellite.
Using the given altitude, the value of g at this height is approximately 0.224 m/s^2. Now, to find the weight of a 2000 kg satellite in a geosynchronous orbit, we use the formula W=mg where m is the mass of the satellite and g is the gravitational acceleration at the altitude. That gives us the value of roughly 448 N for the weight of the satellite at geosynchronous orbit.
Learn more about Geosynchronous Orbit here:https://brainly.com/question/13013840
#SPJ11
A person with mass mp = 71 kg stands on a spinning platform disk with a radius of R = 1.62 m and mass md = 193 kg. The disk is initially spinning at ω = 1.8 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.54 m from the center). 1)What is the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk?
Answer:
439.587 kgm²
Explanation:
[tex]m_p[/tex] = Mass of person = 71 kg
R = Radius of platform = 1.62 m
[tex]m_d[/tex] = Mass of disc = 193 kg
The moment of inertia of the system is given by
[tex]I=m_pR^2+\frac{1}{2}m_dR^2\\\Rightarrow I=71\times 1.62^2+\frac{1}{2}\times 193\times 1.62^2\\\Rightarrow I=439.587\ kgm^2[/tex]
The total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk is 439.587 kgm²
A 66-kg base runner begins his slide into second base when he is moving at a speed of 3.4 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base(a) How much mechanical energy is lost due to friction acting on the runner. (b) How far does he slide?
Answer:
a ) = 381.48 J
b )= 84.25 cm
Explanation:
Kinetic energy of the runner
= 1/2 m v²
= .5 x 66 x 3.4²
= 381.48 J
The final kinetic energy of the runner is zero .
Loss of mechanical energy
= 381.48 J
This loss in mechanical energy is due to action of frictional force .
b )
Let s be the distance of slide
deceleration due to frictional force
= μmg/m
.7 x 66 x 9.8 / 66
a = - 6.86 m s⁻¹
v² = u² - 2 a s
0 = 3.4² - 2x6.86 s
s = 3.4² / 2x6.86
= .8425 m
84.25 cm
Water flows through a valve at the rate of 1000 lbm/s. The pressure just upstream of the valve is 90 psi and the pressure drop across the valve is 50 psi. The inside diameters of the valve inlet and outlet pipes are 12 and 24 in. The flow occurs in a horizontal plane. Determine the loss in available energy units across the valve.
Final answer:
The loss in available energy units across the valve is equal to the pressure drop, which is 344,738 N/m².
Explanation:
To determine the loss in available energy units across the valve, we need to calculate the change in kinetic energy and potential energy before and after the valve. The available energy is given by the equation:
Available energy = pressure drop + change in kinetic energy + change in potential energy
Given that the pressure drop across the valve is 50 psi, we can convert it to SI units (N/m²) by multiplying by 6894.76:
Pressure drop = 50 psi × 6894.76 N/m²/psi = 344,738 N/m²
The change in kinetic energy can be calculated using the following formulas:
Change in kinetic energy = (1/2) × mass flow rate × (outlet velocity² - inlet velocity²)
Change in potential energy = mass flow rate × g × (outlet height - inlet height)
Since the flow occurs in a horizontal plane, there is no change in potential energy. Therefore, the loss in available energy units across the valve is equal to the pressure drop:
Loss in available energy units = 344,738 N/m²
Two kids create a makeshift seesaw by setting a 4-m long uniform plank on a saw horse. The saw horse is 0.5 m to the left of the center of mass of the plank. The child of mass m1 = 42 kg sits at the left end of the plank. The child of mass m2 = 31 kg sits 1 m to the right of the center of mass of the plank. What is the mass of the plank?
Answer:
The mass of the plank is 33 kg.
Explanation:
Given that,
Length of plank = 4 m
Distance in left from the center= 0.5 m
Mass of child = 42 kg
Mass of other child = 31 kg
Distance in right from the center= 1 m
We can sum torques around the saw horse; each force generates a torque equal to force x distance from saw horse
We need to calculate the torque for first child
Using formula of torque
[tex]\tau =mg\times d[/tex]
Put the value into the formula
[tex]\tau=42\times9.8\times1.5[/tex]
[tex]\tau=617.4\ N-m[/tex]
We need to calculate the torque for second child
The child is 1.5 m from the saw horse
Using formula of torque
[tex]\tau' =mg\times d[/tex]
Put the value into the formula
[tex]\tau'=31\times9.8\times1.5[/tex]
[tex]\tau'=455.7\ N-m[/tex]
The weight of the plank produces a clockwise torque
[tex]\tau''=mg\times d[/tex]
[tex]\tau''=mg\times 0.5[/tex]
We need to calculate the mass of the plank
Using balance equation of torque
[tex]\tau=\tau''+\tau'[/tex]
Put the value into the formula
[tex]617.4=mg\times0.5+455.7[/tex]
[tex]m=\dfrac{617.4-455.7}{0.5\times9.8}[/tex]
[tex]m=33\ kg[/tex]
Hence, The mass of the plank is 33 kg.
The mass of the plank is found to be 29 kg.
To solve for the mass of the plank, we need to use the principles of static equilibrium. The seesaw must be balanced, meaning the sum of moments around the pivot (saw horse) must be zero.
Given -
Length of the plank = 4 m
Pivot position (0.5 m to the left of the center) = 2-0.5 m from the left end = 1.5 m
Mass of the first child, m1 = 42 kg (at the left end, 1.5 m from pivot)
Mass of the second child, m2 = 31 kg (1 m to the right of the center, 1.5+1 m = 2.5 m from the pivot)
Equation of moments about the pivot (taking counterclockwise as positive):
Moment of m1 = 42 kg * 1.5 m (clockwise) = -42 * 1.5 Nm
Moment of m2 = 31 kg * 2.5 m (counterclockwise) = 31 * 2.5 Nm
Moment of the plank's weight (assuming it acts at its center, which is 0.5 m from the pivot):
Moment of plank = M * 0.5 m (clockwise) = -M * 0.5 Nm
Setting the sum of moments to zero for equilibrium:
31 * 2.5 - 42 * 1.5 - M * 0.5 = 0
Solving for M:
31 * 2.5 - 42 * 1.5 = M * 0.5
77.5 - 63 = 0.5M
14.5 = 0.5M
M = 14.5 / 0.5 = 29 kg
Therefore, the mass of the plank is 29 kg.
Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?
a. Both hit the ground at the same time.
b. The one thrown at an angle hits the ground with a higher speed.
c. Both hit the ground with the same speed.
d. The one thrown at an angle hits the ground with a lower speed.
Answer:
c. Both hit the ground with the same speed.
Explanation:
Given that speed of the both balls are same.
We know that kinetic energy given as
[tex]KE=\dfrac{1}{2}mv^2[/tex]
v=Speed , m= mass
If mass of the the balls are same then their kinetic energy will be same.
The potential energy PE
PE= m g h
h =height
Given that h is same ,therefore their potential energy will be same.
But the time taken by both the balls will be different.
Therefore answer is c.
Both balls hit the ground with the same speed because they start with the same initial speed and experience the same influence of gravity. Hence, option (c) is the correct answer.
This question pertains to the principles of projectile motion. When two balls are thrown from the same height with the same initial speed, one straight up and one at a 45° angle, we can analyze their motion by breaking it into vertical and horizontal components.
Both hit the ground with the same speed: This is because both balls start with the same initial speed and will have the same kinetic energy upon hitting the ground. The ball thrown straight up will decelerate until it momentarily stops and then accelerates back downward due to gravity. The ball thrown at a 45° angle, however, will have its speed distributed between vertical and horizontal components. Ignoring air resistance, the vertical motion of both balls is symmetrical.The ball thrown straight up will return to its starting height with the same speed it was launched with (but in the opposite direction); combined with the acceleration due to gravity, it will have the same kinetic energy when it strikes the ground as the ball thrown at an angle.Thus, the correct statement is (c) Both hit the ground with the same speed.
A spherical steel ball bearing has a diameter of 2.540 cm at 26.00°C. (Assume the coefficient of linear expansion for steel is 11 ✕ 10−6 (°C)−1. )
(a) What is its diameter when its temperature is raised to 91.0°C? (Give your answer to at least four significant figures.) 165.1 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm
(b) What temperature change is required to increase its volume by 1.100%
Answer: a)2.542cm
Explanation:
According to area expansivity which is defined as change in area per unit area for degree rise in kelvin.
Area expansivity= A2-A1/A1(¶2-¶1)
A2-A1 is change in area
¶2-¶1 is temperature change
A2 if final area
A1 is initial area
¶2 is final temp = 91°C
¶1 is initial temp= 26°C
coefficient of linear expansion for steel is 11 ✕ 10−6 (°C)−1.
Area of the spherical steel ball = Πd²/4
A1= Π×2.54²/4
A1 = 5.07cm²
Area expansivity = 2×linear expansion = 2×11 ✕ 10−6 (°C)−1.
= 22 ✕ 10−6 (°C)−1.
Substituting in the formula to get final area A2
22 ✕ 10−6 (°C)−1 = A2-5.07/5.07(91-26)
22 ✕ 10−6 (°C)−1 = A2-5.07/329.55
A2-5.07 = 0.0073
A2 = 0.0073+5.07
A2= 5.0073cm²
To get final diameter
A2=Πd²/4
5.0073=Πd²/4
20.309 = Πd²
d² = 20.309/Π
d²=6.46
d= √6.46
d= 2.542cm
Two skiers travel two different frictionless paths. The two paths start at the same place and finish at the same place. Skier A is heavier than Skier B. With no more information, which of the skiers has the greater speed at the finish? Ignore air resistance.
A. Skier A has the same speed as Skier B at the finish.
B. Skier A has the greater speed at the finish.
C. Skier B has the greater speed at the finish.
D. There is no way to know which skier has the greater speed at the finish.
Answer:
D. There is no way to know which skier has the greater speed at the finish.
Explanation::
Total energy is conserved because there is no friction
(Et)₀ = (Et)f
(Et)₀ = Initial total energy (J)
(Et)f = Final total energy (J)
(Et)₀= K₀ + U₀
(Et)f= Kf + Uf
K₀ : Initial kinetic energy
U₀ : Initial potential energy
Kf : final kinetic energy
Uf : final potential energy
The formulas to calculate the kinetic energy (K) and potential energy (U) are:
K = ( 1/2)m*v²
U = m* g*h
m : mass (kg)
v: speed ( m/s)
h: hight ( m)
Problem development
Skier A
(Et)₀ = (Et)f
K₀ + U₀ = Kf + Uf
( 1/2)mA*(v₀A)² + mA*g*h₀ = ( 1/2)mA*(vfA)² + mA*g*hf ,
We divide by mA on both sides of the equation
( 1/2)*(v₀A)²+ g*h₀ = ( 1/2)(vfA)² + g*hf
( 1/2)*(v₀A)²+ g*h₀ - g*hf = ( 1/2)(vfA)²
We multiply by 2 both sides of the equation
(v₀A)²+2g(h₀ -hf) = (vfA)²
(vfA)² = (v₀A)²+2g(h₀ -hf) Equation (1)
Skier B
(Et)₀ = (Et)f
K₀ + U₀ = Kf + Uf
(1/2)mB*(v₀B)² + mB*g*h₀ = ( 1/2)mB*(vfB)² + mB*g*hf
We perform the same procedure above:
(vfB)² = (v₀B)²+2g(h₀ -hf) Equation (2)
Comparison of equation (1) with equation (2)
The term 2g (h₀ -hf) is the same in both equations because the paths of the two skiers start in the same place and end in the same place.
The final speed (vf) of skiers depends on their initial speed (v₀).
Because the initial speed of the skiers is unknown, it cannot be determined which has the highest final speed
In Rutherford's scattering experiments, alpha particles (charge =+2e ) were fired at a gold foil. Consider an alpha particle with an initial kinetic energy heading directly for the nucleus of a gold atom (charge =+79e). The alpha particle will come to rest when all its initial kinetic energy has been converted to electrical potential energy.Find the distance of closest approach between the alpha particle and the gold nucleus for the case K= 2.9MeV in fm.
To find the distance of closest approach between the alpha particle and the gold nucleus, set the potential energy equal to zero and solve for r.
Explanation:In Rutherford's scattering experiments, alpha particles with a charge of +2e were fired at a gold foil. To find the distance of closest approach between the alpha particle and the gold nucleus, we need to calculate the potential energy at this point. Given that the initial kinetic energy of the alpha particle is 2.9 MeV, we can equate it to the electrical potential energy using the formula:
KE = PE
Since the alpha particle comes to rest when all its initial kinetic energy has been converted to electrical potential energy, we can set the potential energy equal to zero:
0 = k * (2e * 79e) / r
where k is the Coulomb constant and r is the distance of closest approach.
Learn more about Distance of closest approach here:https://brainly.com/question/31994795
#SPJ6
Final answer:
To determine the distance of closest approach between an alpha particle and a gold nucleus with an initial kinetic energy of 2.9 MeV, we apply the conservation of energy principle and calculate the point where the kinetic energy is entirely converted into electrical potential energy.
Explanation:
The student is asking about the distance of closest approach between an alpha particle and the nucleus of a gold atom during Rutherford's scattering experiments when the initial kinetic energy of the alpha particle is 2.9 MeV. To find this distance, we use the concept of conservation of energy, where the initial kinetic energy of the alpha particle is completely converted into electrical potential energy at the point of closest approach. The formula for the electrical potential energy (U) at distance (r) is given by U = k*q1*q2/r, where k is Coulomb's constant, q1 and q2 are the charges of the alpha particle and gold nucleus, respectively, and r is the distance of closest approach.
The charge of an alpha particle is +2e (+2e), and the charge of a gold nucleus is +79e. Given that the initial kinetic energy (K) is 2.9 MeV, we can set K = U to solve for r. The calculation involves converting the given energy into Joules (1eV = 1.6x10^-19 J), and then using the values for k, q1 (+2e), and q2 (+79e) to determine r. Through this, we can derive the equation to calculate the distance of closest approach in femtometers (fm), noting that 1 fm = 1x10^-15 m.
The y component of the electric field of an electromagnetic wave traveling in the +x direction through vacuum obeys the equation Ey = (375 N/C) cos [kx - (2.20 × 1014 rad/s)t]. What is the wavelength of this electromagnetic wave?
Answer: 8.6 µm
Explanation:
At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:
Ey =Emax cos (kx-ωt)
So, we can write the following equality:
ω= 2.2 1014 rad/sec
The angular frequency and the linear frequency are related as follows:
f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec
In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.
The wavelength, speed and frequency, are related by this equation:
λ = c/f
λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.
The wavelength of the electromagnetic wave is 8.57 micrometers.
The wavelength of an electromagnetic wave can be calculated using the following equation:
λ = c / f
where:
λ is the wavelength
c is the speed of light in vacuum (3.0 × 10^8 m/s)
f is the frequency
The frequency of the electromagnetic wave can be calculated from the angular frequency ω using the following equation:
f = ω / 2π
where:
ω is the angular frequency (2.20 × 10^14 rad/s)
Plugging in the values for ω and c, we get:
f = (2.20 × 10^14 rad/s) / (2π)
f = 3.50 × 10^13 Hz
Now we can calculate the wavelength:
λ = (3.0 × 10^8 m/s) / (3.50 × 10^13 Hz)
λ = 8.57 × 10^-6 m
Therefore, the wavelength of the electromagnetic wave is 8.57 micrometers.
for such more question on wavelength
https://brainly.com/question/10728818
#SPJ3
A block with mass m = 0.250 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at x = +0.240 m, its acceleration is ax = -13.0 m/s2 and its velocity is vx = +4.00 m/s.1. What is the spring's force constant k?2. What is the amplitude of the motion?3. What is the maximum speed of the block during its motion?4. What is the maximum magnitude of the block's acceleration during its motion?
Answer:
13.54 N/m
0.6 m
4.37 m/s
32.496 m/s²
Explanation:
m = Mass of block = 0.25 kg
k = Spring constant
A = Amplitude
x = Compression of spring = 0.24 m
a = Acceleration = -13 m/s²
v = Velocity = 4 m/s
The weight of the block and force on spring is equal
[tex]ma=-kx\\\Rightarrow k=-\frac{ma}{x}\\\Rightarrow k=-\frac{0.25\times -13}{0.24}\\\Rightarrow k=13.54\ N/m[/tex]
The spring's force constant is 13.54 N/m
Total energy of the system is given by
[tex]E=\frac{1}{2}(mv^2+kx^2)\\\Rightarrow E=\frac{1}{2}(0.25\times 4^2+13.54\times 0.24^2)\\\Rightarrow E=2.39\ J[/tex]
At maximum displacement v = 0
[tex]E=\frac{1}{2}(mv^2+kA^2)\\\Rightarrow E=\frac{1}{2}(0+kA^2)[/tex]
[tex]\\\Rightarrow E=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{E2}{k}}\\\Rightarrow A=\sqrt{\frac{2\times 2.39}{13.54}}\\\Rightarrow A=0.6\ m[/tex]
The amplitude of the motion is 0.6 m
Speed of the block
[tex]E=\frac{1}{2}mv_m^2+0\\\Rightarrow v_m=\sqrt{\frac{2E}{m}}\\\Rightarrow v_m=\sqrt{\frac{2\times 2.39}{0.25}}\\\Rightarrow v_m=4.37\ m/s[/tex]
The maximum speed of the block during its motion is 4.37 m/s
Forces in the spring
[tex]ma_m=kA\\\Rightarrow a_m=\frac{kA}{m}\\\Rightarrow a_m=\frac{13.54\times 0.6}{0.25}\\\Rightarrow a_m=32.496\ m/s^2[/tex]
Maximum magnitude of the block's acceleration during its motion is 32.496 m/s²
The force constant of the spring is 13.54 N/m. The amplitude of the oscillatory motion is 0.240 m. The maximum velocity and acceleration are 4.72 m/s and 13 m/s² respectively.
Explanation:First, we can calculate the spring's force constant k using Newton's second law (F = ma) and Hooke's Law (F = kx), by equating the forces F (Note that the negative acceleration is due to the force acting opposite to displacement). Hence, having the force due to spring F = m * a = 0.250 kg * -13.0 m/s² = -3.25 N and applying force due to spring as per Hooke's law as -k * x. Equating these forces gives k = 13.54 N/m.
Secondly, the amplitude A of the motion can be identified as |x| = 0.240 m, since for a simple harmonic oscillator, the amplitude is the maximum displacement from equilibrium (x = 0).
For the third part, the maximum speed Vmax during its motion can be calculated using the energy conservation principle, as given by the formula, Vmax = sqrt(k/m * A²), which gives Vmax = 4.72 m/s.
Finally, the maximum acceleration of the block Amax happens at the points of maximum displacement, i.e., at the amplitude points, which is Amax = k/m * A = 13m/s².
Learn more about Simple Harmonic Motion here:https://brainly.com/question/28208332
#SPJ3