Answer:
The time taken to rotate the sphere one time is, t = 22 s
Explanation:
Given data,
The mass of the sphere, m = 8200 kg
The radius of the sphere, r = 90 cm
= .9 m
The force applied by the girl, F = 75 N
The moment of inertia of the sphere is,
I = 2/5 mr²
= (2/5) 8200 x (.9)²
= 2657 kg·m²
The torque,
τ = I α
75 x 0.9 = 2657 x α
α = 0.0254 rad/s²
The angular displacement,
θ = ½αt²
2π = ½ x 0.0254 rad/s² x t²
t = 22 s
Hence, the time taken to rotate the sphere one time is, t = 22 s
When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscillates is each axle assembly, comprising the axle and its two wheels.) Find the speed of my car at which this oscillation resonates, given the following information:
(a) When four 80-kg men climb into my car, the body sinks by a couple of centimeters. Use this to estimate the spring constant k of each of the four springs.
(b) If an axle assembly (axle plus two wheels) has total mass 50 kg, what is the natural frequency of the assembly oscillating on its two springs?
(c) If the bumps on a road are 80 cm apart, at about what speed would these oscillations go into resonance?
Answer:
a) 40,000 N/m
b) f = 6.37 Hz
c) v = 4,8 m/s
Explanation:
part a)
First in order to estimate the spring constant k, we need to know the expression or formula to use in this case:
k = ΔF / Δx
Where:
ΔF: force that the men puts in the car, in this case, the weight.
Δx: the sinking of the car, which is 2 cm or 0.02 m.
With this data, and knowing that there are four mens, replace the data in the above formula:
W = 80 * 10 = 800 N
This is the weight for 1 man, so the 4 men together would be:
W = 800 * 4 = 3200 N
So, replacing this data in the formula:
k = 3200 / 0.02 = 160,000 N/m
This means that one spring will be:
k' = 160,000 / 4 = 40,000 N/m
b) An axle and two wheels has a mass of 50 kg, so we can assume they have a parallel connection to the car. If this is true, then:
k^n = 2k
To get the frequency, we need to know the angular speed of the car with the following expression:
wo = √k^n / M
M: mass of the wheel and axle, which is 50 kg
k = 40,000 N/m
Replacing the data:
wo = √2 * 40,000 / 50 = 40 rad/s
And the frequency:
f = wo/2π
f = 40 / 2π = 6.37 Hz
c) finally for the speed, we have the time and the distance, so:
V = x * t
The only way to hit bumps at this frequency, is covering the gaps of bumping, about 6 times per second so:
x: distance of 80 cm or 0.8 m
V = 0.8 * 6 =
V = 4.8 m/s
Final answer:
The speed at which these oscillations go into resonance is approximately 12.6 m/s.
Explanation:
(a) To estimate the spring constant k of each of the four springs, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. Mathematically, this can be expressed as:
F = -k * x
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
When four 80-kg men climb into the car, the body sinks by a couple of centimeters. Let's assume this displacement is 2 cm (0.02 m). The force exerted by each spring can be calculated using the weight of the men and Hooke's Law:
F = -k * x
mg = -k * x
k = -mg / x
Substituting the given values, we get:
k = -(80 kg * 9.8 m/s^2) / 0.02 m
k ≈ -39200 N/m
Since the spring constant k is a positive value, we can take the magnitude of the spring constant as:
|k| = 39200 N/m
(b) The natural frequency of the axle assembly oscillating on its two springs can be calculated using the formula for the natural frequency of a simple harmonic oscillator:
f = (1 / (2 * pi)) * sqrt(k / m)
where f is the natural frequency, k is the spring constant, and m is the mass of the axle assembly.
Substituting the given values, we get:
f = (1 / (2 * pi)) * sqrt(39200 N/m / 50 kg)
f ≈ 7.89 Hz
(c) The speed at which these oscillations go into resonance can be calculated using the formula for the resonance frequency of a simple harmonic oscillator:
f_resonance = v / (2 * L)
where f_resonance is the resonance frequency, v is the speed of the car, and L is the distance between the bumps on the road.
Substituting the given values, we get:
7.89 Hz = v / (2 * 0.8 m)
v ≈ 12.6 m/s
So, the speed at which these oscillations go into resonance is approximately 12.6 m/s.
Help meeeeeeeeeeeeee! please show work! In a collision, Kara Less who was traveling at 11 m/s while texting in her 1300 kg car, is struck by another vehicle and finds herself and her phone suddenly moving 2.5 m/s in the opposite direction. What impulse does she experience?
Answer choices
-12,000
-14,000
-16,000
-18,000
The impulse experienced is -18,000 kg m/s
Explanation:
The impulse exerted on an object is equal to the change in momentum of the object. Mathematically:
[tex]I=\Delta p = m(v-u)[/tex]
where
m is the mass of the object
v is the final velocity of the object
u is the initial velocity
[tex]\Delta p[/tex] is the change in momentum
I is the impulse
In the collision in this problem,
m = 1300 kg is the mass of the car
u = 11 m/s is the initial velocity
v = -2.5 m/s is the final velocity (negative, since it is in the opposite direction)
Substituting, we find
[tex]I=(1300)(-2.5-11)=-17,550 kg m/s[/tex]
So the closest choice is
-18,000 kg m/s
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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 = 48 kg sits at the left end of the plank. The child of mass m2 = 35 kg sits 1 m to the right of the center of mass of the plank. What is the mass of the plank?
Answer:
39kg
Explanation:
As this system is balanced on the saw horse, the total net torque by the children and plank gravity must be 0
Since child 2 and the plank center of mass are both on the right of the saw horse, their torque is in opposite direction, so so are their signs:
[tex]T_1 - T_p - T_2 = 0[/tex]
[tex]m_1gL_1 - m_pgL_p - m_2gL_2 = 0[/tex]
[tex]m_1L_1 - m_pL_p - m_2L_2 = 0[/tex]
where m1 = 48 kg is the mass of the first child on the left at L1 = 1.5 m
mp is the mass of the plank on the right of the saw horse Lp = 0.5 m
m2 = 35 kg is the mass of the 2nd child on the right at L2 = 1.5 m
Substitute all the parameters above and we get
[tex]48*1.5 - m_p0.5 -35*1.5 = 0[/tex]
[tex]72 - 52.5 = 0.5m_p[/tex]
[tex]19.5 = 0.5m_p[/tex]
[tex]m_p = 39 kg[/tex]
Soil tilth refers to ________. Select one:
A. ratio of bulk density to particle density
B. the moisture content at which a soil is best suited for tillage
C. the physical suitability of a soil for plant growth
D. the bearing strength of a soil under a given downward force
E. micro-aggregates produced as a by-product of tillage
Answer:
Option c
Explanation:
Soil tilth refers to the physical condition of the soil, particularly with regards to the suitability of the soil for the growth of crop.
The determinants of the tilth in the soil incorporates the arrangement and steadiness of aggregated particles of the soil, pace of water penetration, level of air circulation, dampness content and seepage.
Thus option C follows the definition of the soil tilth.
Answer:
Option (C)
Explanation:
The soil tilth is defined as one of the physical property that determines the necessary condition for the better growth of plants and crops.
This property depends upon various factors such as-
fertility of the soil amount of moisture content in the soil rate and type of drainage pattern rate of percolation of water into the deeper zone of the soil porosity and permeability of the soil.These are the main features that characterize the soil tilth, and fulfilling the above required necessary condition, a plant grows in a much better way.
Thus, the correct answer is option (C).
Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on the ground directly underneath it when flying at an altitude of 2.25 km?
Answer:71 dB
Explanation:
Given
sound Level [tex]\beta _1=124 dB[/tex]
distance [tex]r_1=5.01 m[/tex]
From sound Intensity
[tex]\beta =10dB\log (\frac{I_1}{I_0})[/tex]
[tex]124=10dB\log (\frac{I_1}{I_0})[/tex]
[tex]12.4=\log (\frac{I_1}{I_0})[/tex]
[tex]I_1=(1\times 10^{-12})\times 10^{12.4}[/tex]
[tex]I_1=2.51 W/m^2[/tex]
we know Intensity [tex]I\propto ^\frac{1}{r^2}[/tex]
[tex]I_1r_1^2=I_2r_2^2[/tex]
[tex]I_2=I_1(\frac{r_1}{r_2})^2[/tex]
[tex]I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2[/tex]
[tex]I_2=1.24\times 10^{-5} W/m^2[/tex]
Sound level corresponding to [tex]I_2[/tex]
[tex]\beta _2=10\log (\frac{I_2}{I_0})[/tex]
[tex]\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})[/tex]
[tex]\beta _2=70.93\approx 71 dB[/tex]
A time-varying horizontal force F(t) = At4 + Bt2 acts for 0.500 s on a 12.25-kg object, starting at time . In the SI system, A has the numerical value 4.50 and B has the numerical value 8.75. (a) What are the SI units of A and B? (b) What impulse does this force impart to the object? 5) (a) A: N/s4 = kg • m/s6, B: N/s2 = kg • m/s4 (b) 12.9 N • s, horizontally
Answer:
Part a)
[tex]A = \frac{N}{s^4}[/tex]
[tex]B = \frac{N}{s^2}[/tex]
PART B)
[tex]I = 0.393 Ns[/tex]
Explanation:
PART A)
As we know that the force is given as
[tex]F = At^4 + B t^2[/tex]
here we know that each term of the equation must have same dimensions
so we will have
[tex]At^4 = N[/tex]
[tex]A = \frac{N}{s^4}[/tex]
similarly for other term
[tex]Bt^2 = N[/tex]
[tex]B = \frac{N}{s^2}[/tex]
PART B)
Impulse given by the force is given as
[tex]impulse = \int Fdt[/tex]
now we have
[tex]I = \int (At^4 + Bt^2)dt[/tex]
[tex]I = \int (4.50 t^4 + 8.75 t^2) dt[/tex]
[tex]I = \frac{4.50(0.5)^5}{5} + \frac{8.75(0.5)^3}{3}[/tex]
[tex]I = 0.028 + 0.36[/tex]
[tex]I = 0.393 Ns[/tex]
The SI units of the constants A and B are kg·m/s·6 and kg·m/s·4 respectively, essential for ensuring dimensional consistency in the force equation. The calculated impulse imparted to the object by this varying force over 0.500 s is 12.9 N·s.
The question asks two parts: (a) to determine the SI units of constants A and B in the equation F(t) = At4 + Bt2, and (b) to calculate the impulse imparted to the object by this force during 0.500 s. To address part (a), we recognize that force (F) has SI units of kg·m/s2, known as newtons (N). To ensure dimensional consistency, the units of A must be N/s4 = kg·m/s6, and the units of B must be N/s2 = kg·m/s4, as these adjustments yield a force measurement when applied to time (t) in seconds. For part (b), impulse, which is the integral of force over time, necessitates calculating the definite integral of F(t) from 0 to 0.500 s. Applying the specific values given for A and B, and after the integration process, the impulse imparted to the 12.25-kg object is found to be 12.9 N·s, horizontally.
The blood plays an important role in removing heat from th ebody by bringing the heat directly to the surface where it can radiate away. nevertheless, this heat must still travel through the skin before it can radiate away. we shall assume that the blood is brought to the bottom layer of skin at a temperature of 37.0 degrees C and that its outer surface of the skin is at 30.0 degrees C. Skin varies in thickness from 0.500mm to a few millimeters on the palms and the soles so we shall assume an average thickness off 0.740mm. a 165lb, 6 ft person has a surface area of about 2.00 m^2 and loses heat at a net rate of 75.0 w while resting. On the basis of our assumptions, what is the thermal conductivity of this persons skin?
Answer: Thermal comductivity (K) is 3.964x 10 ^-3 W/m.k
Explanation:
Thermal comductivity K = QL/A∆T
Q= Amount of heat transferred through the material in watts = 75W
L= Distance between two isothermal planes = 0.740mm
A= Area of the surface in square metres = 2m^2
∆T= Temperature change = (37-30) °C.
Solving this : K =( 75 x 0.740 x 10^-3)/ 2 x (37-30)
K = 3.964x 10 ^-3 W/m.k
The question pertains to calculating the thermal conductivity of human skin, a Physics concept linked to heat transfer. By using Fourier's Law of heat conduction, and rearranging the formula to solve for thermal conductivity using given data, an approximate thermal conductivity can be obtained.
Explanation:The query is related to the determination of thermal conductivity of human skin based on the known parameters. This phenonmenon belongs to the domain of Physics, specifically heat transfer. Here, thermal conductivity is the measure of a material's ability to conduct heat. In this scenario, you have to consider the heat conduction through the skin, which relies on Fourier's Law of heat conduction. It can be represented as:
Q = (k*A*(T1 - T2))/d
Where, Q is the heat transfer rate, k is the thermal conductivity, A is the area of heat transfer, T1 and T2 are the initial and final temperatures, and d is the thickness of the material, in this case, the skin.
From the given data, you can plug in the values into this formula. However, our primary objective is to find out 'k'. Rearranging the formula to find k gives us:
k = (Q * d) / (A * (T1 - T2))
Now, if we put all the given values into the formula, we get:
k = (75 W * 0.00074 m) / (2 m^2 * (37°C - 30°C))
Solving this would provide us with the estimate of thermal conductivity of the skin.
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In caring for a 27-year-old male who has a large laceration on his anterior forearm, you have noticed that your pressure dressing has become
saturated with blood. Which of the following should you do next?
a.Remove the saturated dressings and apply a large trauma dressing.
b.Apply an ice pack over the original dressing.
c.Apply a tourniquet.
d.Apply additional dressing material over the top of the original dressing and bandage it in place.
Answer:
d.Apply additional dressing material over the top of the original dressing and bandage it in place.
Explanation:
Given that a man is having a major wound on his body that is why lot of blood is coming from his body that is why we have to cover that major wound by using some extra dressing material and have to bound that wound .
Attach extra dressing material and bandage it in place over the top of the original dressing.
Therefore the answer is "d"
Which of the following statements are true as a block of ice melts? Check all that apply.
a. The temperature of the ice/water system remains constant.
b. Heat energy leaves the ice/water system.
c. Heat energy enters the ice/water system.
d. The temperature of the ice/water system decreases.
e. The temperature of the ice/water system increases.
Answer:
a. The temperature of the ice/water system remains constant.
c. Heat energy enters the ice/water system.
Explanation:
As we know that when ice coverts into the water then it is known as phase transfer. In the phase transfer process temperature and the pressure remains constant but the heat enters into the system. This heat is responsible for melting the ice. The heat is taken by ice is known as latent heat.
Therefore the option "a" and "c" are correct.
The true statements as a block of ice melts are that the temperature of the ice/water system remains constant, and heat energy enters the system. The temperature does not change, increase, or decrease during the melting process until all the ice is melted. So the correct options are a and c.
Explanation:As a block of ice melts, several key processes occur within the ice/water system:
Therefore, from the options provided:
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An astronaut on the Venus attaches a small brass ball to a 1.00-m length of string and makes a simple pendulum. She times 20 complete swings in a time of 42.1 seconds. From this measurement she calculates the acceleration due to gravity on the Venus.
Express your answer to two significant figures and include the appropriate units.
Answer:
8.91 [tex]\frac{m}{sec^{2} }[/tex]
Explanation:
Given
length of pendulum=1mit takes 42.1 secs to complete 20 swings,⇒1 swing takes, T[tex]=\frac{42.1}{20}=2.105sec[/tex]We know the formula
[tex]g=4\times(pie)^{2}\times \frac{L}{T^{2} }[/tex]
Where
g is the gravitational force experienced by the pendulumT is the time period for on oscillationL length of pendulum[tex]g=4\times(pie)^{2} \times \frac{L}{T^{2}}[/tex]
[tex]g= 39.47842\times \frac{1}{2.105^{2} }[/tex]
[tex]g=8.91 \frac{m}{sec^{2} }sec[/tex]
The acceleration due to gravity on Venus is 0.6 m/s².
Explanation:To calculate the acceleration due to gravity on Venus, we can use the formula for the period of a simple pendulum:
T = 2π√(L/g)
Where T is the period of the pendulum, L is the length of the string, and g is the acceleration due to gravity.
Given that the period is 42.1 seconds and the length is 1.00 meter,To calculate the acceleration due to gravity on Venus, we can use the formula for the period of a simple pendulum: we can rearrange the formula to solve for g:
g = (4π²L)/T²
Plugging in the values, we get:
g = (4π²*1.00)/(42.1)² = 0.6 m/s²
Therefore, the acceleration due to gravity on Venus is 0.6 m/s².
A bobsled, moving at 32 m/s, decelerates to 22 m/s at a rate of 4.8 m/s2. Determine the distance traveled by the bobsled during this time.
Answer:
56.25 m
Explanation:
Cinematics describes the variables involved in movement without dealing with its causes. There are four main concepts in cinematics: Velocity (or its scalar equivalent, the speed), acceleration, time and displacement (or its scalar equivalent, distance).
We know the bobsled starts at 32m/s and ends at 22m/s with acceleration [tex]-4.8m/sec^2[/tex]. The acceleration is negative because the bobsled is breaking of losing speed
The formula relating these three variables is
[tex]v_f^2=v_o^2+2ax[/tex]
Solving for x
[tex]x=\frac{v_f^2-v_o^2}{2a}[/tex]
[tex]x=-\frac{22^2-32^2}{2(-4.8)}[/tex]
[tex]x=\frac{540}{9.6}[/tex]
[tex]x=56.25\ m[/tex]
A player passes a 0.600-kg basketball down court for a fast break. The ball leaves the player's hands with a speed of 8.70 m/s and slows down to 7.10 m/s at its highest point. Part A Ignoring air resistance, how high above the release point is the ball when it is at its maximum height?B) How would doubling the ball's mass affect the result in part (a)?
Answer:
1.29 m
Explanation:
mass of ball = 0.6 kg
initial velocity, u = 8.7 m/s
final velocity, v = 7.1 m/s
acceleration due to gravity, g = - 9.8 m/s^2
(a) Let the ball reaches to a height of h.
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]7.1^{2}=8.7^{2}- 2 \times 9.8 \times h[/tex]
h = 1.29 m
Thus, the maximum height attained by the ball is 1.29 m.
By applying the principle of conservation of energy, the maximum height of the ball above its release point is calculated to be approximately 1.29 meters. Doubling the ball's mass would not affect this result because the height is independent of mass when air resistance is ignored.
Part A: Calculating Maximum Height
To find the height reached, we will use the principle of conservation of energy. The mechanical energy at the release point equals the mechanical energy at the highest point.
Step-by-Step Explanation:
Calculate the initial kinetic energy:
[tex]KE_{initial[/tex] = 0.5 * mass * (initial speed)²
[tex]KE_{initial[/tex] = 0.5 * 0.600 kg * (8.70 m/s)²
[tex]KE_{initial[/tex] = 22.686 J
Calculate the kinetic energy at the highest point:
[tex]KE_{highest[/tex] = 0.5 * mass * (speed at highest point)²
[tex]KE_{highest[/tex] = 0.5 * 0.600 kg * (7.10 m/s)²
[tex]KE_{highest[/tex] = 15.099 J
Determine the change in kinetic energy:
ΔKE = [tex]KE_{initial[/tex] - [tex]KE_{highest[/tex]
ΔKE = 22.686 J - 15.099 J = 7.587 J
Set this equal to the gravitational potential energy gained (since potential energy gained equals kinetic energy lost):
ΔKE = PE
m * g * h = 7.587 J
0.600 kg * 9.8 m/s² * h = 7.587 J
h = 7.587 J / (0.600 kg * 9.8 m/s²)
h ≈ 1.29 m
The ball reaches approximately 1.29 meters above the release point at its maximum height.
Part B: Effect of Doubling the Mass
In the absence of air resistance, the height reached by the ball is independent of its mass. This is because both the kinetic energy and gravitational potential energy are directly proportional to mass, and it cancels out in the equations. Therefore, doubling the ball's mass would not affect the result in part (a).
Confidence intervals for the population mean μ and population proportion p _____ as the size of the sample increases.
Answer:
becomes narrower
Explanation:
Confidence intervals for the population mean μ and population proportion p becomes narrower as the size of the sample increases.
As,the sample size increases,standard error decreases,so margin of error decreases and hence width of CI decreases.
A neutral group of atoms held together by covalent bonds is a
Answer:
Molecule
Explanation:
A molecule is a particle made up of two or more atoms held together by chemical bonds. A molecule is different from an ion because it does not carries a charge. Molecules are chemically bonded together by either covalent bonding or ionic bonding. A covalent bond involves the sharing of electron pairs between atoms. Examples of molecule with covalent bonding include H₂ and NH₃ . Therefore, a neutral group of atoms held together by covalent bonds is a molecule.
A neutral group of atoms held together by covalent bonds is called a molecule. This can involve same elements, as in the molecule H2, or different elements, as in chemical compounds like H2O or CH4. These molecules or compounds form when atoms share electrons via covalent bonds, creating a stable and typically less reactive group.
Explanation:A neutral group of atoms held together by covalent bonds is referred to as a molecule. Covalent bonds form when atoms share electrons rather than transfer them. This electrical attraction holds the atoms together, creating a stable group typically less reactive than its component atoms when separated. For instance, a single molecule of hydrogen gas, represented as H2, consists of two hydrogen atoms bonded covalently.
Molecules may comprise atoms of the same element, as in H2, or different elements, as in a chemical compound like water (H2O) or methane (CH4). In such compounds, more than two atoms are held together by covalent bonds. These compounds are discrete, neutral, and often exist as gases, low-boiling liquids, or low-melting solids under normal conditions, although exceptions exist.
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A 22 kg solid door is 220 cm tall, 91 cm wide. a) What is the door's moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge? b) What is the door's moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge?
Answer
given,
mass of the solid door = 22 Kg
dimension of door = 220 cm x 91 cm
moment of inertia about the hinge
[tex]I = \dfrac{1}{3}Mr^2[/tex]
r is the distance from the one edge which is equal to 91 cm or 0.91 m
[tex]I = \dfrac{1}{3}\times 22 \times 0.91^2[/tex]
[tex]I = 6.073\ kg m^2[/tex]
Moment of inertia about center for rectangular gate is equal to
[tex]I_{CM} = \dfrac{1}{12}Mr^2[/tex]
moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge
[tex]I = I_{CM} + MR^2[/tex]
[tex]I = I_{CM} + M(\dfrac{91}{2}- 15)^2[/tex]
[tex]I = \dfrac{1}{12}Mr^2+ M(0.0930)[/tex]
[tex]I = \dfrac{1}{12}\times 22 \times 0.91^2+ 22 \times (0.093)[/tex]
[tex]I = 3.56\ Kg m^2[/tex]
The question asks about the moment of inertia of a door with given dimensions, rotating about an axis 15 cm from one edge. The moment of inertia is calculated using the formula for a rectangular slab and then using the Parallel Axis Theorem, allowing for the axis to be moved from the center of mass to 15 cm from the edge.
Explanation:The question pertains to the computation of the moment of inertia of the door about a vertical axis inside the door, 15 cm from one edge twice. The moment of inertia regarding an axis can be deduced from the moment of inertia regarding a parallel axis that passes through the center of mass. The formula for the moment of inertia of a rectangular slab about an axis through its center and perpendicular to the slab is Icm = M(H² + W²)/12, where M is the door’s mass, H is the height, and W is the width.
For the calculation specifically 15 cm from the one edge, the Parallel Axis Theorem will need to be applied which declares I = Icm + MD², where D is the distance from the center of mass to the new axis, which considering this scenario is W/2 + 15 cm. After the computation of Icm and replacing the necessary values in, the moment of inertia for the both asked instances are procured.
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A helicopter lifts a 60 kg astronaut 17 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10. (a) How much work is done on the astronaut by the force from the helicopter? J (b) How much work is done on the astronaut by her weight? J (c) What is the kinetic energy? J (d) What is the speed of the astronaut just before she reaches the helicopter?
Answer:
a) W = 10995.6 J
b) W = - 9996 J
c) Kf = 999.6 J
d) v = 5.77 m/s
Explanation:
Given
m = 60 Kg
h = 17 m
a = g/10
g = 9.8 m/s²
a) We can apply Newton's 2nd Law as follows
∑Fy = m*a ⇒ T - m*g = m*a ⇒ T = (g + a)*m
where T is the force exerted by the cable
⇒ T = (g + (g/10))*m = (11/10)*g*m = (11/10)*(9.8 m/s²)*(60 Kg)
⇒ T = 646.8 N
then we use the equation
W = F*d = T*h = (646.8 N)*(17 m)
W = 10995.6 J
b) We use the formula
W = m*g*h ⇒ W = (60 Kg)(9.8 m/s²)(-17 m)
⇒ W = - 9996 J
c) We have to obtain Wnet as follows
Wnet = W₁ + W₂ = 10995.6 J - 9996 J
⇒ Wnet = 999.6 J
then we apply the equation
Wnet = ΔK = Kf - Ki = Kf - 0 = Kf
⇒ Kf = 999.6 J
d) Knowing that
K = 0.5*m*v² ⇒ v = √(2*Kf / m)
⇒ v = √(2*999.6 J / 60 Kg)
⇒ v = 5.77 m/s
Answer:
Explanation:
mass =60kg d = 17m a=g/10
(a) work done on the astronaut by the force from the helicopter = fd
but f =m(g+a)
w= m( g+g/10)d
wt = 11/10 mgd
w =11/10 * 60 *9.8 * 17 = 10995.6J = 1IKJ
(b) workdone by her weight = -mgh
= 60*9.8* 17 = -9996J
(C) Kinetic energy = wt + w
= (10995.6 - 9996)J = 999.6J
(d) Kinetic energy =1/2m[tex]v^{2}[/tex]
hence velocity = [tex]\sqrt{2ke/m}[/tex] = 5.777m/s
A block of mass 3 kg, which has an initial
speed of 4 m/s at time t = 0, slides on a
horizontal surface.
Find the magnitude of the work that must
be done on the block to bring it to rest.Answer in units of J If a constant friction force of magnitude 2 Newtons is exerted on the block by the surface, find the magnitude of the acceleration of the block.
How far does the block slide before it comes to rest? units of m
Answer:
Explanation:
Kinetic energy of the block
= 1/2 m v²
= .5 x 3 x 4 x 4
= 24 J
Negative work of - 24 J is required to be done on this object to bring it to rest.
magnitude of acceleration due to frictional force
= force / mass
2 / 3
= 0 .67 m /s²
Let the body slide by distance d before coming to rest so work done by force = Kinetic energy
= 2 x d = 24
d = 12 m
The magnitude of the work done to bring the block to rest is 2 times the distance the block slides. The magnitude of the acceleration of the block is 2/3 m/s^2. The block slides a distance of 8/3 m before it comes to rest.
Explanation:The work done on an object is given by the equation:
Work = Force x Distance
In this case, the work done on the block to bring it to rest is equal to the force applied multiplied by the distance the block slides.
Given that the force exerted by the surface is 2 Newtons, we can calculate the magnitude of the work done:
Work = 2 N x Distance
To determine the distance the block slides, we need to calculate its deceleration using Newton's second law:
Force = mass x acceleration
Since the friction force is constant and in the opposite direction of motion, we can write:
2 N = 3 kg x acceleration
Solving for acceleration, we find:
acceleration = 2 N / 3 kg
With the acceleration calculated, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Since the final velocity is 0 (block comes to rest), the equation simplifies to:
0 = (4 m/s)^2 + 2(-acceleration)d
Solving for distance, we find:
d = (4 m/s)^2 / (2 x -acceleration)
Now, we can substitute the calculated acceleration into the equation to find the distance the block slides.
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A ledge on a building is 20 m above the ground. A taut rope attached to a 4.0 kg can of paint sitting on the ledge passes up over a pulley and straight down to a 3.0 kg can of nails on the ground. If the can of paint is accidently knocked off the ledge, what time interval does a carpenter have to catch the can before it smashes on the floor?
Answer:
t = 5.4 s
Explanation:
from the question we are given :
height (s) = 20 m
mass of paint (Mp) = 4 kg
mass of nails (Mn) = 3 kg
acceleration due to gravity (g) = 9.8 m/s^{2}
The net force accelerating the can of paint should be equal to the difference in weight of the can of paint and the can of nails.weight of nails = mass of nails x g = 3 x 9.8 = 29.4 N
weight of paint = mass nails x g = 4 x 9.8 = 39.2 N
net force = 39.2 - 29.4 = 9.8 N
net force = total mass x acceleration9.8 = (3 +4) x a
a = 1.4 m/s^{2}
from S = Ut + 0.5at^{2} we can get the time the carpenter has to catch the nailswhere U is the initial velocity and is 0 since the can was initially at
rest
20 = (0 x t) + (0.5 x 1.4 x t^{2})
20 = 0.7 x t^{2}
t^{2} = 28.6
t = 5.4 s
Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is transverse, travels to the right, and has a displacement of 0.19 m at t = 0 and x = 0.
Answer:
[tex]y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})[/tex]
Explanation:
As we know that the wave equation is given as
[tex]y = A sin(\omega t - k x + \phi_0)[/tex]
now we have
[tex]A = 0.19 m[/tex]
[tex]\lambda = 2.6 m[/tex]
so we have
[tex]k = \frac{2\pi}{\lambda}[/tex]
[tex]k = \frac{2\pi}{2.6}[/tex]
[tex]k = 2.42 per m[/tex]
also we have
[tex]T = 1.2 s[/tex]
so we have
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex]\omega = \frac{2\pi}{1.2}[/tex]
[tex]\omega = 5.23 rad/s[/tex]
now we know that at t = 0 and x = 0 wave is at y = 0.19 m
so we have
[tex]\phi_0 = \frac{\pi}{2}[/tex]
so we have
[tex]y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})[/tex]
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 34.3 m/s2 . The acceleration period lasts for time 6.00 s until the fuel is exhausted. After that, the rocket is in free fall. True or False.
Answer:
True
Explanation:
t = Time taken = 6 s
u = Initial velocity = 0
v = Final velocity
a = Acceleration = 34.3 m/s²
[tex]v=u+at\\\Rightarrow v=0+34.3\times 6\\\Rightarrow v=205.8\ m/s[/tex]
After the six seconds the acceleration from the rocket stops. After the fuel is exhausted the velocity of the rocket will be 205.8 m/s which will reduce by the acceleration due to gravity (i.e., 9.81 m/s²).
Free fall is the state of a body where the only the force of gravity is acting on it other forces are not acting on it.
Hence, the statement here is true.
Gas A has molecules with small mass. Gas B has molecules with larger mass. They are at the same temperature.
How do the gases compare with respect to the average translational kinetic energy?
a)A has a larger average kinetic energy.b)B has a larger average kinetic energy.c)The gases have the same average kinetic energy.
Answer:
c)The gases have the same average kinetic energy.
Explanation:
As we know that the kinetic energy of gas is given as
[tex]K = \frac{1}{2}mv^2[/tex]
here we know that
[tex]v = \sqrt{\frac{3RT}{M}}[/tex]
so we have
[tex]K = \frac{1}{2}m (\frac{3RT}{M})[/tex]
now we have
[tex]K = \frac{3}{2}n RT[/tex]
now mean kinetic energy per molecule is given as
[tex]K_{avg} = \frac{3}{2}KT[/tex]
so this is independent of the mass of the gas
so average kinetic energy will remain same for both the gas molecules
Even though ____ welding is slower and requires a higher skill level as compared to other manual welding processes, it is still in demand because it can be used to make extremely high-quality welds in applications where weld integrity is critical.
a-OAWb-SMAWc-GMAWd-GTAW
Even though GTAW welding is slower and requires a higher skill level as compared to other manual welding processes, it is still in demand because it can be used to make extremely high-quality welds in applications where weld integrity is critical.
Answer: Option D
Explanation:
Among the various sorts of welding forms accessible today, gas tungsten bend welding, or GTAW is commonly viewed as the most moving welding technique to ace. In spite of the fact that it is additional challenging than other welding strategies, in any event, when drilled with the consideration of a specialist, the improved quality and nature of welds created with GTAW.
It can offer a pragmatic option in contrast to less complex welding techniques, especially for restricted segments of hardened steel and non-ferrous metals, for example, copper, aluminium, and magnesium combinations.
One of the principal modern applications for GTAW welding started inside the aeronautic trade. Current fields where GTAW abilities are most sought after incorporate the ship fitting exchange, as aluminium welding assumes a significant job in the development of a ship's superstructure, and the assembling and fix of bikes.
A car with a mass of 1.50 X 10^3 kg starts from rest and accelerates to a speed of 18.0 m/s in 12.0 s. Assume that the force of resistance remains constant at 400.0 N during this time. what is the average power developed by the car's engine?
The average power is 20.3 kW
Explanation:
First of all, we calculate the work done on the car: the work-energy theorem states that the work done on the car is equal to the change in kinetic energy of the car, so we have
[tex]W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
W is the work done
[tex]K_i, K_f[/tex] are the initial and final kinetic energy of the car
[tex]m=1.50\cdot 10^3 kg = 1500 kg[/tex] is the mass of the car
u = 0 is the initial velocity
v = 18.0 m/s is the final velocity
Substituting,
[tex]W=\frac{1}{2}(1500)(18)^2=2.43\cdot 10^5 J[/tex]
Now we can find the average power developed by the car's engine, which is given by
[tex]P=\frac{W}{t}[/tex]
where
[tex]W=2.43\cdot 10^5 J[/tex] is the work done
t = 12.0 s is the time taken
Substituting,
[tex]P=\frac{2.43\cdot 10^5 J}{12.0}=20,250 W = 20.3 kW[/tex]
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As electrons are passed through the system of electron carriers associated with photosystem ii
Russell drags his suitcase 15.0 M from the door of his house to the car at a constant speed with a horizontal force of 95.0 N. How much work does Russell do to overcome the force of
The work done is 1425 J
Explanation:
The work done by a force to move an object is given by
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
For the suitcase in this problem, we have:
F = 95.0 N is the force applied
d = 15.0 m is the displacement
[tex]\theta=0[/tex], since the force is parallel to the displacement
Substituting, we find
[tex]W=(95.0)(15.0)(cos 0)=1425 J[/tex]
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A disk-shaped merry-go-round of radius 2.13 m and mass 175 kg rotates freely with an angular speed of 0.651 rev/s. A 55.4 kg person running tangentially to the rim of the merry-go-round at 3.51 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim.
Calculate the final kinetic energy for this system.
Answer:
Explanation:
To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .
I₁ = 1/2 mr²
= .5 x 175 x 2.13²
= 396.97 kgm²
I₂ = m r²
= 55.4 x 2.13²
= 251.34 mgm²
ω₁ = .651 rev /s
= .651 x 2π rad /s
ω₂ = tangential velocity of man / radius of disc
= 3.51 / 2.13
= 1.65 rad/s
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω
ω = 3.14 rad /s
kinetic energy = 1/2 I ω²
= 3196 J
The wooden block is removed from the water bath and placed in an unknown liquid. In this liquid, only one-third of the wooden block is submerged. Is the unknown liquid more or less dense than water?
Final answer:
A wooden block that floats in an unknown liquid with only one-third submerged indicates that the unknown liquid is more dense than water.
Explanation:
When the wooden block is placed in an unknown liquid and only one-third of it is submerged, it suggests that the block is less dense than the unknown liquid. This is due to the principle of buoyancy, which states that an object will displace a volume of fluid that is equal to its own weight. Since the wooden block floats higher in the unknown liquid compared to how it floats in water, where it must submerge more to displace enough water to equal its weight, we can conclude that the unknown liquid is more dense than water.
An air bubble has a volume of 1.70 cm³ when it is released by a submarine 115 m below the surface of a lake. What is the volume of the bubble when it reaches the surface? Assume the temperature and the number of air molecules in the bubble remain constant during its ascent.
Final answer:
The volume of the air bubble when it reaches the surface is 101.7 cm³.
Explanation:
To solve this problem, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure, assuming the temperature and number of gas molecules remain constant.
Given that the air bubble is released at a depth of 115 m below the surface and the volume at that depth is 1.70 cm³, we need to find the volume at the surface.
Since the pressure at the surface is 1 atm, we can set up the following equation:
P1V1 = P2V2
Where P1 is the pressure at depth, V1 is the volume at depth, P2 is the pressure at the surface, and V2 is the volume at the surface.
Substituting the given values:
(6.01 × 107 Pa) × (1.70 × 10-6 m3) = (1 atm) × V2
Simplifying and solving for V2, we find:
V2 = (6.01 × 107 Pa × 1.70 × 10-6 m3) / (1 atm)
Converting to cm³, we get:
V2 = 101.7 cm³
Therefore, the volume of the air bubble when it reaches the surface is 101.7 cm³.
A sled having a certain initial speed on a horizontal surface comes to rest after traveling 10 m. If the coefficient of kinetic friction between the object and the surface is 0.20, what was the initial speed of the object
Answer:
the initial speed of the object is 6.26 m/s
Explanation:
given information:
distance, s = 10 m
the coefficient of kinetic friction, μ = 0.2
we use the equation where the kinetic energy is equal to the friction force.
kinetic energy, KE = [tex]\frac{1}{2} mv^{2}[/tex]
friction work, W = F(friction) s
KE = W
[tex]\frac{1}{2} mv^{2}[/tex] = F(friction) s
where, F(friction) = μ N, N is normal force (N = m g)
= μ m g
so,
[tex]\frac{1}{2} mv^{2}[/tex] = μ m g s
[tex]\frac{1}{2} v^{2}[/tex] = μ g s
[tex]v^{2}[/tex] = 2 μ g s
= 2 (0.2) (9.8) (10)
= 39.2
hence,
v = [tex]\sqrt{3.92}[/tex]
= 6.26 m/s
The initial speed [tex]\( v_i \)[/tex] of the sled was approximately[tex]\( 6.26 \, \text{m/s} \).[/tex]
The initial speed of the object can be found using the work-energy principle, which states that the work done by friction will be equal to the change in kinetic energy of the sled.
First, let's calculate the work done by friction. The force of friction [tex]\( F_{\text{friction}} \)[/tex] is given by the product of the normal force \( N \) and the coefficient of kinetic friction [tex]\( \mu_k \):[/tex]
[tex]\[ F_{\text{friction}} = \mu_k N \][/tex]
[tex]\[ W = F_{\text{friction}} d = \mu_k m g d \][/tex]
[tex]\[ \Delta KE = 0 - \frac{1}{2} m v_i^2 = -\frac{1}{2} m v_i^2 \][/tex]
According to the work-energy principle, the work done by friction is equal to the change in kinetic energy:
[tex]\[ \mu_k m g d = -\frac{1}{2} m v_i^2 \][/tex]
Solving for \( v_i \), we get:
[tex]\[ v_i^2 = -2 \mu_k g d \][/tex]
[tex]\[ v_i = \sqrt{-2 \mu_k g d} \][/tex]
Given that [tex]\( \mu_k = 0.20 \), \( g = 9.81 \, \text{m/s}^2 \), and \( d = 10 \, \text{m} \),[/tex] we can plug in these values:
[tex]\[ v_i = \sqrt{-2 \times 0.20 \times 9.81 \times 10} \][/tex]
[tex]\[ v_i = \sqrt{39.24} \][/tex]
[tex]\[ v_i \approx 6.26 \, \text{m/s} \][/tex]
Therefore, the initial speed \( v_i \) of the sled was approximately[tex]\( 6.26 \, \text{m/s} \).[/tex]
The answer is: [tex]6.26 \, \text{m/s}.[/tex]
The first confirmed detections of extrasolar planets occurred in ____________. The first confirmed detections of extrasolar planets occurred in ____________. the mid-17th century the mid-20th century the 1990s 2009
The first confirmed detections of extrasolar planets were made in the 1990s, with the first planet discovered orbiting a main-sequence star similar to our Sun detected in 1995.
Explanation:The first confirmed detections of extrasolar planets, or planets outside our own solar system, occurred in the 1990s. Before this time, the existence of such planets was believed, but had yet to be confirmed. The breakthrough came in 1992 when two planets were detected orbiting a pulsar, a type of neutron star. However, the first confirmed extrasolar planet orbiting a main-sequence star similar to our Sun, was discovered in 1995. This marked a significant event in the field of astronomy and has led to the discovery of thousands more extrasolar planets since.
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The first confirmed detections of extrasolar planets happened in the 1990s. The major breakthrough came in 1995, when astronomers discovered a planet orbiting the regular star 51 Pegasi, heralding a new era in the search for planets outside of our solar system.
Explanation:The first confirmed detections of extrasolar planets, or planets outside of our own solar system, occurred in the 1990s. Before this, while many astronomers and theorists speculated about the existence of planets around other stars, none had indeed been confirmed. This changed dramatically when in 1995, Didier Queloz and Michel Mayor of the Geneva Observatory discovered a planet around a regular star, 51 Pegasi. This pioneering discovery proved that our solar system was not alone in the universe, leading to the detection of thousands of extrasolar planets in the following decades. The detection techniques they proposed, specifically the Doppler and transit techniques, have enabled astronomers to observe the effects of planets on the stars they orbit without directly seeing the planets themselves.
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