Answer:
Radius of the outer most dark fringe is 2.65 cm
Solution:
As per the question:
Radius of curvature of the glass, r = 10.8 m
No. of dark fringes, n = 100
Wavelength of light, [tex]\lambda = 652\ nm = 652\times 10^{- 9}\ m[/tex]
Now,
To calculate the radius R of the outermost ring:
Radius of the dark fringe of nth order is given by:
[tex]R^{2} = nr\lambda = 100\times 10.8\times 652\times 10^{- 9} = 7.042\times 10^{- 4}[/tex]
[tex]R = \sqrt{7.042\times 10^{- 4}} = 0.0265\ m = 2.65\ cm[/tex]
Part complete Horizontally-polarized light passes through a polarizing sheet, and only 25 % of the intensity of the incident light is transmitted through the sheet. What angle does the polarization of the light make with the horizontal after passing through the polarizer?
Answer:
60°
Explanation:
I₀ = Intensity of unpolarized light
θ = Angle between the axis of the filter and polarization direction
Intensity of polarzied light
[tex]I=I_0cos\theta[/tex]
Here, the light that is transmitted is reduced by 25% that means
[tex]I=0.25I_0[/tex]
So,
[tex]0.25I_0=I_0cos^2\theta\\\Rightarrow cos^2\theta =0.25\\\Rightarrow cos\theta =5\\\Rightarrow \theta= cos^{-1}0.5\\\Rightarrow \theta=60^{\circ}[/tex]
∴ The angle between the axes of the polarizer and the analyzer is 60°
A window in a house has a rectangular shape of 2.0 m by 1.0 m. The glass in the window is 0.5 cm thick, with a thermal conductivity of 0.84 W/(m oC). The temperature in the house is 20.5 °C, and the temperature outside is 4 oC. How much heat is lost through this window in one minute? Don't use scientific notations
Answer:
P = 5280 W
Explanation:
The conductivity of the materials determines that heat flows from the hot part to the cold part, the equation for thermal conductivity transfer is
P = Q / t = k A ([tex]T_{h}[/tex] -[tex]T_{c}[/tex]) / L
Where k is the thermal conductivity of the glass 0.8 W / ºC, A the area of the window, T the temperature and L is glass thickness
Let's calculate the window area
A = l * a
A = 2.0 1.0
A = 2.0 m²
Let's replace
L = 0.5 cm (1 m / 100 cm) = 0.005 m
P = 0.8 2 (20.5 - 4) / 0.005
P = 5280 W
For a particular reaction, Δ H ∘ = − 93.8 kJ and Δ S ∘ = − 156.1 J/K. Assuming these values change very little with temperature, at what temperature does the reaction change from nonspontaneous to spontaneous?
To solve this problem it is necessary to apply the concepts related to Gibbs free energy and spontaneity
At constant temperature and pressure, the change in Gibbs free energy is defined as
[tex]\Delta G = \Delta H - T\Delta S[/tex]
Where,
H = Entalpy
T = Temperature
S = Entropy
When the temperature is less than that number it is negative meaning it is a spontaneous reaction. [tex]\Delta G[/tex] is also always 0 when using single element reactions. In numerical that implies [tex]\Delta G = 0[/tex]
At the equation then,
[tex]\Delta G = \Delta H - T\Delta S[/tex]
[tex]0 = \Delta H - T\Delta S[/tex]
[tex]\Delta H = T\Delta S[/tex]
[tex]T = \frac{\Delta H}{\Delta S}[/tex]
[tex]T = \frac{-93.8kJ}{-156.1J/K}[/tex]
[tex]T = \frac{-93.8*10^3J}{-156.1J/K}[/tex]
[tex]T = 600.89K[/tex]}
Therefore the temperature changes the reaction from non-spontaneous to spontaneous is 600.89K
A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.
What is the relation between the standard free Gibbs energy and the spontaneity of a reaction?When ΔG° < 0, the reaction is spontaneous.When ΔG° = 0, the reaction is not spontaneous nor nonspontaneous.When ΔG° > 0, the reaction is nonspontaneous.We want to know at what temperature a reaction changes from nonspontaneous to spontaneous, that is, at what temperature ΔG° = 0.
Given the standard enthalpy and entropy of the reaction, we can calculate that temperature using the following expression.
ΔG° = ΔH° - T . ΔS°
0 = -93.8 kJ - T . (-156.1 J/K)
T = 601 K
A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.
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A railroad tank car contains milk and rolls at a constant speed along a level track. The milk begins to leak out the bottom. The car then
a. Need more information about the rate of the leak.
b. maintains a constant speed.
c. speeds up.
d. slows down.
The answer is option d. slows down.
Explanation:The answer to the question is option d. slows down.
When the milk begins to leak out of the bottom of the railroad tank car, the car will experience a decrease in its mass as the milk is being lost. According to the Law of Conservation of Momentum, if the mass of an object decreases without any external forces acting on it, the object will slow down.
Therefore, as the milk leaks out of the tank car, its speed will slow down over time.
A block of mass m attached to a horizontally mounted spring with spring constant k undergoes simple harmonic motion on a rictionless surface. How would the maximum speed of the block be affected i the spring constant was increased by a factor of 4 while holding the amplitude of oscillation constant?
1) It would remain unchanged.
2) It would increase by a factor of 2.
3) It would decrease by a factor of 1/4
4) It would increase by a factor of 4.
5) It would decrease by a factor of 1/2
To solve this problem it is necessary to apply the concepts related to simple harmonic movement.
The maximum speed from the simple harmonic motion is given as
[tex]V = A\sqrt{\frac{K}{m}}[/tex]
Where,
K = Spring constant
m = mass
At this case m is a constant then
[tex]V \propto \sqrt{K}[/tex]
then the ratio is given by
[tex]\frac{v_2}{v_1}=\sqrt{\frac{K_1}{K_2}}[/tex]
According the statement,
[tex]v_2 = \sqrt{\frac{K_1}{K_2}}v_1[/tex]
[tex]v_2 = 2v_1[/tex]
Therefore the maximum speed becomes double: 2) It would increase by a factor of 2.
Two identical billiard balls traveling at the same speed have a head-on collision and rebound. If the balls had twice the mass, but maintained the same size and speed, how would the rebound be different?
a. No difference.
b. At a higher speed.
c. At slower speed.
Doubling the mass of billiard balls during an elastic head-on collision would result in no difference in the rebound speed due to the conservation of momentum. The speed and motion of the balls post-collision would be unaffected by the change in mass.
Explanation:The question asks how the rebound of two identical billiard balls would be different if their mass were doubled, but they maintained the same size and speed during a head-on collision. According to the principle of conservation of momentum, which dictates that the total momentum of an isolated system remains constant if no external forces act upon it, the rebound should be unaffected by the change in mass.
In a head-on elastic collision, if we assume an ideal scenario without energy loss due to factors like friction or air resistance, doubling the mass of the balls while keeping the speed the same would not change the speed at which they rebound. This is because the momentum before the collision must equal the momentum after the collision for each ball, and since momentum is the product of mass and velocity (p = mv), the velocities would remain unchanged post-collision. Therefore, the correct answer would be (a) No difference.
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A ring made from brass has an inner radius of 2.50000 cm and an outer radius of 3.50000 cm, giving the ring a thickness of 1.00000 cm. The thermal expansion coefficient of brass is 0.000019/°C. If the temperature of the ring is increased from 20.0°C to 90.0°C, by how much does the thickness of the ring change? _______ cm
Answer:0.00133 cm
Explanation:
Given
change in temperature [tex]\Delta T=90-20=70^{\circ}C[/tex]
Thermal Expansion coefficient of brass [tex]\alpha =0.000019 /^{\circ}C[/tex]
thickness t=1 cm
change in thickness is given by
[tex]\Delta t=t\cdot \alpha \cdot \Delta T[/tex]
[tex]\Delta t=1\times 0.000019\times 70[/tex]
[tex]\Delta t=0.00133\ cm[/tex]
Starting with an initial value of P(0) = 75, the population of a prairie dog community grows at a rate of
P(t) =50 − t/5 (in units of prairie dogs/month), for 0 <_t <_250.
a. What is the population 99 months later?
b. Find the population P(t) for 0 <_t <_250.
Two castings are proposed for a study. One casting is a cube 10.16 cm (4 in.) on a side; the other is a rectangle 10.64 cm (16 in.) tall × 10.16 cm (4 in.) wide × 2.45 cm (1 in.) thick. Both are cast from the same metal at the same temperature, and the mold material is the same in both cases. If the cube-shaped casting solidifies in 14.5 min, how rapidly do we expect the rectangular casting to solidify?
To solve the problem it is necessary to apply the concepts related to Chvorinov's Law, which states that
[tex]\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2[/tex]
Where,
[tex]V_c[/tex] = Volume cube
[tex]SA_c[/tex] = Superficial Area from Cube
[tex]V_r[/tex] = Volume Rectangle
[tex]SA_r[/tex]= Superficial Area from Rectangle
Our values are given as (I will try to develop the problem in English units for ease of calculations),
[tex]V_c = 4^3 = 64in^3[/tex]
[tex]SA_c = 6*4^2=96in^2[/tex]
[tex]SA_r = 2*(1*16+4*1+16*4)=168in^2[/tex]
[tex]V_r = 4*1*16 = 64in^3[/tex]
Applying the Chvorinov equation we have to,
[tex]\frac{T_r}{T_c} = (\frac{V_r}{V_c}*\frac{SA_c}{SA_r})^2[/tex]
[tex]\frac{T_r}{T_c} = (\frac{64}{64}*\frac{96}{168})^2[/tex]
[tex]\frac{T_r}{T_c} = (\frac{96}{168})^2[/tex]
[tex]\frac{T_r}{T_c} = 0.3265[/tex]
The stipulated time for the cube is 14.5 then,
[tex]T_r = 0.3265*14.5[/tex]
[tex]T_r = 4.735min[/tex]
You push very hard on a heavy desk, trying to move it. You do work on the desk:
a. only if it doesn't move.
b. whether or not it moves, as long as you are exerting a force.
c. only if it starts moving.
d. never—it does work on you.
Answer:
(C) Only if it starts moving
Explanation:
We know that work done is given by
[tex]W=F.d=Fdcos\Theta[/tex]
So there are two case in which work done is zero
First case is that when force and displacement are perpendicular to each other
And other case is that when there is no displacement
So for work to be done there must have displacement, if there is no displacement then there is no work done
So option (c) will be the correct option
Work in a physical sense is done on a desk only if the desk moves when you push it. Simply exerting force is not considered work without displacement in the direction of the force. So the correct option is C.
Explanation:When considering whether you do work on a heavy desk by pushing it, it's important to understand the scientific definition of work. In physics, work is defined as the transfer of energy that occurs when a force is applied over a distance. Therefore, you do work on an object if, and only if, the object moves in the direction of the force. So the correct answer to the question is:
c. only if it starts moving.
Simply exerting a force on an object does not constitute work unless the object is displaced. For example, if you continue to push against a wall and it does not move, despite your effort and energy consumption, physically no work is done because there is no displacement.
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Two boats leave a dock together . Each travels in a straight line . The angle between their courses measures 54° 10¢ . One boat travels 36.2 km/hr and the other travels 45.6 km/hr. How far apart will they be after three hours
Answer:
218.93 km
Explanation:
[tex]\theta[/tex] = Angle between the paths of the two boats = 54.10°
Distance = Speed × Time
Distance traveled by one boat = [tex]36.2\times 3\ km[/tex]
Distance traveled by other boat = [tex]45.6\times 3\ km[/tex]
From the triangle law of vectors we have
[tex]d=\sqrt{(36.2\times 3)^2+(45.6\times 3)^2+2\times 36.2\times 3\times 45.6\times 3\times cos54.10}\\\Rightarrow d=218.93\ km[/tex]
The distance they will be apart after 3 hours is 218.93 km
A machine carries a 33.6 kg package from an initial position of r0 = (0.502 + 0.751 + 0.207) m at t0 = 0 s to a final position of r1 = (7.82 + 2.17 + 7.44) m at t1 = 11.9 s. The constant force applied by the machine on the package is F = (21.5 + 42.5 + 63.5) N.
Answer:
The work done on the package by the machine's force is 676.94 J.
Explanation:
Given that,
Mass of package = 33.6 kg
Initial position [tex]r_{0}=(0.502i+0.751j+0.207k)\ m[/tex]
Final position [tex]r_{1}=(7.82i+2.17j+7.44k)\ m[/tex]
Final time = 11.9 s
Force [tex]F=(21.5i+42.5j+63.5k)[/tex]
Suppose we need to find the work done on the package by the machine's force
We need to calculate the displacement
Using formula of displacement
[tex]d=r_{1}-r_{0}[/tex]
Put the value into the formula
[tex]d=(7.82i+2.17j+7.44k)-(0.502i+0.751j+0.207k)[/tex]
[tex]d=7.318i+1.419j+7.233k[/tex]
We need to calculate the work done
Using formula of work done
[tex]W=F\dotc d[/tex]
[tex]W=(21.5i+42.5j+63.5k)\dotc(7.318i+1.419j+7.233k)[/tex]
[tex]W=676.94\ J[/tex]
Hence, The work done on the package by the machine's force is 676.94 J.
A frictionless piston-cylinder device contains 4.5 kg of nitrogen at 110 kPa and 200 K. Nitrogen is now compressed slowly according to the relation PV1.5 = constant until it reaches a final temperature of 360 K. Calculate the work input during the process, in kJ.
Answer:
427.392 kJ
Explanation:
m = Mass of gas = 4.5 kg
Initial temperature = 200 K
Final temperature = 360 K
R = Mass specific gas constant = 296.8 J/kgK
[tex]\gamma[/tex] = Specific heat ratio = 1.5
Work done for a polytropic process is given by
[tex]W=\frac{mR\Delta T}{1-\gamma}\\\Rightarrow W=\frac{4.5\times 296.8(360-200)}{1-1.5}\\\Rightarrow W=-427392\ J\\\Rightarrow W=-427.392\ kJ[/tex]
The work input during the process is -427.392 kJ
Find a glass jar with a screw-top metal lid. Close the lid snugly and put the jar into the refrigerator. Leave it there for about 10 minutes and then take the jar out and try to open the lid. (a) Did the lid become tighter or looser? Explain your observation.
Answer:
The lid becomes tighter
It becomes tighter because metals have a lower heat capacity than glass meaning their temperature drops (or increases) much faster than glass for the same energy change. So in this example, the metal will contract faster than the glass causing it to be more tighter around the glass.
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6640 N (1493 lbf). If the length of the rod is 370 mm (14.57 in.), what must be the diameter to allow an elongation of 0.53 mm (0.02087 in.)?
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as
[tex]\Delta L=\dfrac{PL}{AE}[/tex]
[tex]A=\dfrac{PL}{\Delta LE}[/tex]
[tex]A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}[/tex]
A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm
Answer:
[tex]d=7.32\ mm[/tex]
Explanation:
Given:
Young's modulus, [tex]E=110\times 10^3\ MPa[/tex]yield strength, [tex]\sigma_y=240\ MPa[/tex]load applied, [tex]F=6640\ N[/tex]initial length of rod, [tex]l=370\ mm[/tex]elongation allowed, [tex]\Delta l=0.53[/tex]We know,
Stress:
[tex]\sigma=\frac{F}{A}[/tex]
where: A = cross sectional area
Strain:
[tex]\epsilon = \frac{\Delta l}{l}[/tex]
& by Hooke's Law within the elastic limits:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
[tex]\therefore 110\times 10^3=\frac{F}{A}\div \frac{\Delta l}{l}[/tex]
[tex]\therefore 110\times 10^3=\frac{6640\times 4}{\pi.d^2}\div \frac{0.53}{370}[/tex]
where: d = diameter of the copper rod
[tex]d=7.32\ mm[/tex]
The production of sound during speech or singing is a complicated process. Let's concentrate on the mouth. A typical depth for the human mouth is about 8.00 cm, although this number can vary. (Check it against your own mouth.) We can model the mouth as an organ pipe that is open at the back of the throat. A) What are the wavelengths of the first four harmonics you can produce if your mouth is open? (in cm/4 answers) B)What are the frequencies of the first four harmonics you can produce if your mouth is open? Use 354 (in Hz/4 answers) C)What are the wavelengths of the first four harmonics you can produce if your mouth is closed? (in cm/4 answers) D)What are the frequencies of the first four harmonics you can produce if your mouth is closed?
Answer:A) The wavelengths are:8cm for first harmonic
5.3cm for second harmonic
4cm for third harmonic
3.2cm for fourth harmonic
B) The frequencies are:
44.25Hz for first harmonic
66.38Hz for second harmonic
88.5Hz for third harmonic
110.63Hz for fourth harmonic
C) The wavelengths are:
10.66Cm for first harmonic
6.4 cm for second harmonic
4.57cm for third harmonic
3.5cm for fourth harmonic
D) The frequencies are:
33.19Hz for first harmonic
55.31 Hz for second harmonic
77.44Hz for third harmonic
99.56Hz for fourth harmonic
Explanation: since the mouth is being modelled as an organ pipe, then we can say that an open mouth is a model of an organ pipe open at both ends.
For an open pipe all harmonics are possible but only odd harmonic's are possible in closed pipes. A closed mouth is a model of a closed pipe, since closed pipes are open at one end.
19.A 20 kg sign is pulled by a horizontal force such that the single rope (originally vertical) holding the sign makes an angle of 21° with the vertical. Assuming the sign is motionless, findA) the magnitude of the tension in the rope andB) the magnitude of the horizontal force.
Answer:
A)[tex]T=209.94N[/tex]
B) [tex]F=75.24N[/tex]
Explanation:
Using the free body diagram and according to Newton's first law, we have:
[tex]\sum F_y=Tcos(21^\circ)-mg=0(1)\\\sum F_x=F-Tsin(21^\circ)=0(2)[/tex]
A) Solving (1) for T:
[tex]T=\frac{mg}{cos(21^\circ)}\\T=\frac{20kg(9.8\frac{m}{s^2})}{cos(21^\circ)}\\T=209.94N[/tex]
B) Solving (2) for F:
[tex]F=Tsin(21^\circ)\\F=(209.94N)sin(21^\circ)\\F=75.24N[/tex]
The energy in an oscillating LC circuit containing a 1.57 H inductor is 5.76 μJ. The maximum charge on the capacitor is 201 μC. For a mechanical system with the same period, find the
(a) mass
(b) spring constant
(c) maximum displacement
(d) maximum speed.
Answer:
(a)1.57 kg
(b) 281.17 N/m
(c) 201 micrometer
(d) [tex]2.69\times 10^{-3}m/sec[/tex]
Explanation:
We have given that value of inductor L = 1.57 Henry
Inductive energy [tex]E=5.76\mu j=5.76\times 10^{-6}J[/tex]
Maximum charge [tex]Q=201\mu C=201\times 10^{-6}C[/tex]
(A) In electrical mechanical system mass corresponds to inductance
So mass will be m = 1.57 kg
(B) We have given energy [tex]E=\frac{Q^2}{2C}[/tex]
[tex]C=\frac{Q^2}{2E}=\frac{(201\times 10^{-6})^2}{2\times 5.7\times 10^{-6}}=3543.94\times 10^{-6}[/tex]
In electrical mechanical system spring constant is equivalent to [tex]\frac{1}{C}[/tex]
So spring constant [tex]k=\frac{1}{C}=\frac{1}{3543.94\times 10^{-6}}=282.17N/m[/tex]
(c) Displacement is equivalent to maximum charge
So displacement will be [tex]x=201\mu m[/tex]
(d) Maximum speed is correspond to maximum current
As maximum current [tex]i_m=\frac{Q}{\sqrt{LC}}=\frac{201\times 10^{-6}}{\sqrt{1.57\times 3543.94\times 10^{-6}}}=2.69\times 10^{-3}A=2.69\times 10^{-3}m/sec[/tex]
A typical meteor that hits the earth’s upper atmosphere has a mass of only 2.5 g, about the same as a penny, but it is moving at an impressive 40 km/s. As the meteor slows, the resulting thermal energy makes a glowing streak across the sky, a shooting star. The small mass packs a surprising punch. At what speed would a 900 kg compact car need to move to have the same kinetic energy
Answer:u=66.67 m/s
Explanation:
Given
mass of meteor [tex]m=2.5 gm\approx 2.5\times 10^{-3} kg[/tex]
velocity of meteor [tex]v=40km/s \approx 40000 m/s[/tex]
Kinetic Energy of Meteor
[tex]K.E.=\frac{mv^2}{2}[/tex]
[tex]K.E.=\frac{2.5\times 10^{-3}\times (4000)^2}{2}[/tex]
[tex]K.E.=2\times 10^6 J[/tex]
Kinetic Energy of Car
[tex]=\frac{1}{2}\times Mu^2[/tex]
[tex]=\frac{1}{2}\times 900\times u^2[/tex]
[tex]\frac{1}{2}\times 900\times u^2=2\times 10^6 [/tex]
[tex]900\times u^2=4\times 10^6[/tex]
[tex]u^2=\frac{4}{9}\times 10^4[/tex]
[tex]u=\frac{2}{3}\times 10^2[/tex]
[tex]u=66.67 m/s[/tex]
Answer:
v = 67 m/s
Explanation:
The meteor has a mass (m) of 2.5 g and a speed (v) of 40 km/s. In SI units:
2.5 g × (1 kg / 10³ g) = 2.5 × 10⁻³ kg
40 km/s × (10³ m / 1 km) = 4.0 × 10⁴ m/s
The kinetic energy (KE) is:
KE = 1/2 × m × v² = 1/2 × (2.5 × 10⁻³ kg) × (4.0 × 10⁴ m/s)² = 2.0 × 10⁶ J
A 900 kg compact car, with the same kinetic energy, must have the following speed.
KE = 1/2 × m × v²
2.0 × 10⁶ J = 1/2 × 900 kg × v²
v = 67 m/s
Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?
The measurement will be significantly affected.
Recall that the relationship between linear velocity and angular velocity is subject to the formula
[tex]v = \omega r[/tex],
Where r indicates the radius and [tex]\omega[/tex] the angular velocity.
As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.
Mounting larger-diameter tires on a truck leads to a systematic error in the speedometer reading, whereby the speedometer will underreport the truck's true linear speed due to an increase in linear distance covered per tire rotation at a constant angular velocity.
If larger-diameter tires are mounted on a truck, the speedometer reading will not be correct. This is due to the relationship between the angular velocity of the tires and the linear speed of the truck. Since the speedometer measures the angular speed of the tires to determine the truck's linear speed, changing to larger tires will cause a systematic error in the reading. For a fixed angular velocity, larger tires will cover a greater linear distance because of their larger circumference. Thus, the speedometer will display a speed lower than the truck's true linear speed. This affects the accuracy of the speedometer.
To further illustrate, consider a truck tire rotating with an angular velocity α. The linear (tangential) velocity v at the surface of the tire is given by v = rα, where r is the tire radius. With larger tires, for the same angular velocity α, r is bigger, hence v is larger. Therefore, when the tires are larger, the speedometer, which is calibrated for the standard tire size, underestimates the truck's speed because it is based on an incorrect assumption about the tire circumference.
Spiderman, whose mass is 70.0 kg, is dangling on the free end of a 12.2-m-long rope, the other end of which is fixed to a tree limb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to swing enough that he can reach a ledge when the rope makes a θ = 58.4° angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver?
Answer:
U = -3978.8 J
Explanation:
The work of the gravitational force U just depends of the heigth and is calculated as:
U = -mgh
Where m is the mass, g is the gravitational acceleration and h the alture.
for calculate the alture we will use the following equation:
h = L-Lcos(θ)
Where L is the large of the rope and θ is the angle.
Replacing data:
h = 12.2-12.2cos(58.4)
h = 5.8 m
Finally U is equal to:
U = -70(9.8)(5.8)
U = -3,978.8 J
A 2.00 kg, frictionless block s attached to an ideal spring with force constant 550 N/m. At t = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 10.0 m/s. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Angular frequency, frequency, and period in shm.find:-a) the amplitude.b) the maximum acceleration of the block.c) the maximum force the spring exerts on the block.
The amplitude of the motion is 0.6 m. The maximum acceleration of the block is 165 m/s^2. The maximum force exerted by the spring on the block is 330 N.
Explanation:The problem you're asking about is related to Simple Harmonic Motion (SHM) associated with a spring-block system on a frictionless surface. An ideal spring obeys Hooke's law which states that the force it exerts is proportional to the displacement from its equilibrium position.
Amplitude of the motion can be determined using the principle of conservation of energy. In the absence of frictional forces, the kinetic energy of the block at maximum displacement converts entirely into potential energy of the spring. So, (1/2)mv^2 = (1/2)kA^2, where m is the mass of the block, v is the velocity of the block, k is the spring constant and A is the amplitude. Substituting the given values, we have A = sqrt((2*2*10^2)/550) = 0.6 m. The maximum acceleration a of the block happens at the maximum displacement, i.e., at the amplitude. According to Hooke's law, F = -kx, where F is the force exerted by the spring and x is the displacement. So, we have a = F/m = kA/m = 550*0.6/2 = 165 m/s^2. The maximum force the spring exerts on the block is when the block is at the amplitude. It equals F = kA = 550*0.6 = 330 N.Learn more about Simple Harmonic Motion here:
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Calculate the acceleration of gravity on the surface of the Sun. The mass of the Sun is MSun = 1.99 ✕ 1030 kg, the radius of the Sun is rSun = 6.96 ✕ 108 m, and G = 6.67 ✕ 10−11 N · m2/kg2. m/s2 (b) By what factor would your weight increase if you could stand on the Sun? (Never mind that you can't.) Fg, Sun Fg, Earth =
Answer: 28
Explanation:
According to Newton's law of Gravitation the gravitational force on the surface of the Sun [tex]Fg_{s}[/tex] is:
[tex]Fg_{s}=mg_{sun}=G\frac{Mm}{r^{2}}[/tex] (1)
Where:
[tex]m[/tex] is your mass
[tex]g_{sun}[/tex] is the acceleration due gravity on the surface of the Sun
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the gravitational constant
[tex]M=1.99(10)^{30} kg[/tex] is the mass of the Sun
[tex]r=6.96(10)^{8} m[/tex] is the radius of the Sun
Simplifying:
[tex]g_{sun}=G\frac{M}{r^{2}}[/tex] (2)
[tex]g_{sun}=6.674(10)^{-11} \frac{m^{3}}{kgs^{2}} \frac{1.99(10)^{30} kg}{(6.96(10)^{8} m)^{2}}[/tex] (3)
[tex]g_{sun}=274 m/s^{2}[/tex] (4)
Since the acceleration due gravity on Earth is [tex]g_{E}=9.8 m/s^{2}[/tex], the relation is:
[tex]\frac{g_{sun}}{g_{E}}=\frac{274 m/s^{2}}{9.8 m/s^{2}}[/tex]
[tex]\frac{g_{sun}}{g_{E}}=27.9 \approx 28[/tex]
Hence, your weight would increase by a factor of 28.
The acceleration of gravity on the surface of the Sun is approximately 274 m/s^2. If you could stand on the Sun, your weight would increase by a factor of approximately 274.
Explanation:To calculate the acceleration of gravity on the surface of the Sun, we can use the formula for gravitational acceleration: g = GM/R^2. Given that the mass of the Sun (M) is 1.99 ✕ 10^30 kg, the radius of the Sun (R) is 6.96 ✕ 10^8 m, and the gravitational constant (G) is 6.67 ✕ 10^-11 N · m^2/kg^2, we can substitute these values into the formula to find the acceleration of gravity on the surface of the Sun. Using the formula, the acceleration of gravity on the surface of the Sun is approximately 274 m/s^2.
To calculate the factor by which your weight would increase if you could stand on the Sun, we can use the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2 and the weight of a 1.0-kg object is 9.8 N. Since the acceleration of gravity on the surface of the Sun is 274 m/s^2, we can calculate the weight on the Sun by multiplying the mass by the acceleration of gravity on the Sun. Dividing the weight on the Sun by the weight on Earth will give us the factor by which your weight would increase if you could stand on the Sun. Using this calculation, the weight on the Sun would be approximately 274 times greater than the weight on Earth.
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An individual experiences a deep cut that severs the radial artery near the elbow. Ignoring air resistance, approximately how high will the blood spurt? (Hints: the specific gravity of blood is 1.050 g cm-3 and the specific gravity of mercury is 13.6 g cm-3 )
The blood will spurt to a height of 0.077 meters or 7.7 centimeters.
Height of blood spurtThe height to which blood spurts from a severed artery can be calculated using Torricelli's law, which is based on the principles of fluid dynamics.
The formula for the height h that blood spouts from the artery can be derived from Torricelli's law:
h = (2 * ΔP) / (ρ * g),
where:
ΔP is the change in pressure,ρ is the density of the fluid (blood),g is the acceleration due to gravity.To find ΔP, we can calculate the pressure difference between the cut artery and the atmosphere using the following formula:
ΔP = ρ * g * h.
Given that the specific gravity of blood is 1.050 g cm⁻³, we can find its density:
ρ_blood = [tex](1.050 g cm^{-3}) * (1 g / cm^3) = 1.050 g / cm^3.[/tex]
The specific gravity of mercury is 13.6 g[tex]cm^{-3[/tex], which corresponds to its density:
ρ_mercury = 13.6 g / [tex]cm^3[/tex].
Now, we can find the pressure difference:
ΔP = [tex](1.050 g / cm^3) * (9.81 m/s^2) * h[/tex],
where g is the acceleration due to gravity (approximately 9.81 m/s²).
We'll assume that the height (h) is in meters. We want to find h, so we'll solve for it:
h = ΔP / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).
Now, plug in the values:
h = ΔP / ((1.050 g / cm³) * (9.81 m/s²)).
h = (ρ_mercury * g * h) / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).
Now, solve for h:
h = (13.6 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex]) * h / ((1.050 g / [tex]cm^3[/tex]) * (9.81 [tex]m/s^2[/tex])).
h = (13.6 / 1.050) * h.
h = 12.952 * h.
h = h / 12.952.
1 = 1 / 12.952.
h = 1 / 12.952.
h ≈ 0.077 m.
So, the blood will spurt to a height of approximately 0.077 meters or 7.7 centimeters.
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The blood spurt from a deep cut is determined by the blood pressure, which needs to exceed the vein pressure for fluid entry, and by the pressure gradient from the heart to smaller vessels. These factors, combined with the density disparity between blood and mercury, influence the height of the blood spurt.
Explanation:To determine how high the blood will spurt when an individual experiences a deep cut that severs the radial artery near the elbow, we first need to understand the pressure at which the fluid, in this case blood, enters the vascular system. It must exceed the blood pressure in the vein, which is approximately 18 mm Hg above atmospheric pressure.
The blood pressure of a young adult is represented by 120 mm Hg at systolic (maximum output of the heart) and 80 mm Hg at diastolic (due to the elasticity of arteries maintaining pressure between heartbeats). Thus, the density of mercury which raises the mercury fluid in the manometer is 13.6 times greater than water, meaning the height of the fluid will be 1/13.6 of that in a water manometer.
The blood leaves the heart with a pressure of about 120 mm Hg but its pressure continues to decrease as it goes from the aorta to smaller arteries to small veins. The pressure differences in the circulation system are due to the blood flow as well as the position of the person.
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Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field exceeds 0.100 T. In the absence of any external magnetic field, determine the maximum current a 4.50-mm-diameter niobium wire can carry and remain superconducting.
To develop the problem it is necessary to apply the concepts related to Magnetic Field.
The magnetic field is defined as
[tex]B = \frac{\mu_0 I}{2\pi r}[/tex]
Where,
[tex]\mu_0 =[/tex] Permeability constant in free space
r = Radius
I = Current
Our values are given as,
B = 0.1T
d = 4.5mm
r = 2.25mm
If the maximum current that the wire can carry is I, then
[tex]B = \frac{\mu_0 2I}{4\pi r}[/tex]
[tex]I = \frac{Br}{2\frac{\mu_0}{4\pi}}[/tex]
[tex]I = \frac{(0.1T)(2.25*10^{-3}m)}{2(1*10^{-7}N/A^2)}}[/tex]
[tex]I = 1125A[/tex]
Therefore the maximum current is 1125A
If a rectangular area is rotated in a uniform electric field from the position where the maximum electric flux goes through it to an orientation where only half the flux goes through it, what has been the angle of rotation? A) 45° B) 26.6° C) 90° D) 30° E
The correct angle of rotation that results in half the electric flux going through a rectangular area in a uniform electric field is 60 degrees, which corresponds to option D) 30 degrees.
The angle of rotation that results in half the electric flux can be found by understanding the relationship between the angle of rotation and the flux through a surface in a uniform electric field. The electric flux through a surface is calculated using the formula [tex]\\( \Phi _{E} = EA \cos(\theta) \)[/tex], where E is the electric field strength, A is the area of the surface, and ([tex]\theta[/tex]is the angle between the field and the normal to the surface. The maximum flux is when [tex]\\(\theta)[/tex] is 0 degrees. To find the angle at which half of the maximum flux passes through, we set [tex](\cos(\theta))[/tex] equal to 0.5 (since [tex](\cos(\theta)\)[/tex] is the ratio of the current flux to the maximum flux), which corresponds to an angle [tex](\theta[/tex] = 60 degrees. Therefore, if the rectangle initially had maximum flux passing through it, the rotation angle that results in half the flux is 60 degrees, which corresponds to answer D) 30 degrees.
The angle of rotation at which the electric flux through the surface is halved is 60°, as it corresponds to the angle where the cosine of the angle equals 0.5. This corresponds to option D in the multiple-choice question.
The electric flux through a surface is given by the equation ∅ = E*A*cosФ where
E is the magnitude of the electric field, A is the area of the surface, and Ф is the angle between the field and the normal to the surface.For maximum flux, the angle is 0° since (cos(0°) = 1).
If the flux is halved, the equation becomes
Ф/2 = E*A* cosФ
To find the angle of rotation for which the flux is halved, we set (cos(Ф) = 1/2), which corresponds to ( Ф= 60°) or (Ф = 120°).
The angle of rotation from the position of maximum flux (0°) to the position where half the flux goes through it is therefore either 60° or 120°. However, since the question implies a single rotation direction, we would generally take the smaller angle, which is 60°, corresponding to option D.
What is the pressure (in N/m2) inside an alveolus having a radius of 2.22 ✕ 10−4 m if the surface tension of the fluid-lined wall is the same as for soapy water? You may assume the pressure is the same as created by a spherical bubble.
Answer:
The pressure inside the bubble is 666.67 [tex]N/m^{2}[/tex]
Solution:
As per the question:
Radius, R = [tex]2.22\times 10^{- 4}\ m[/tex]
Now,
Given that the surface tension of the wall is the same as that of soapy water.
The air trapped inside the bubble exerts pressure on the soap bubble which is given by:
Gauge Pressure, P = [tex]\frac{4T}{r}[/tex]
Also, the surface tension of the soapy water, [tex]T_{s} = 0.0370\ N/m^{2}[/tex]
To calculate the pressure inside the alveolus:
[tex]P_{i} = \frac{4T}{R} = \frac{4T_{s}}{R}[/tex]
[tex]P_{i} = \frac{4\times 0.0370}{2.22\times 10^{- 4}} = 666.67\ N/m^{2}[/tex]
The New River Gorge Bridge in West Virginia is a steel arch bridge 518 m in length. How much does the total length of the roadway decking change between temperature extremes of –20.0°C and 35.0°C? The result indicates the size of the expansion joints that must be built into the structure.
Final answer:
The change in length of the New River Gorge Bridge due to thermal expansion can be calculated using the coefficient of linear expansion for steel and the temperature change noted.
Explanation:
The question is about calculating the change in length of a steel structure, the New River Gorge Bridge, due to thermal expansion between its temperature extremes. Using a standard linear expansion formula, ΔL = αLδT, where ΔL is the change in length, α is the coefficient of linear expansion for steel (typically around 11×10^-6 1/°C), L is the length of the steel at its coldest, and δT is the change in temperature. For the New River Gorge Bridge with a length of 518 m and a temperature change from -20.0°C to 35.0°C (a change of 55.0°C), the expansion can be calculated as follows: ΔL = (11×10^-6 1/°C)×518 m×55.0°C, which gives the total change in length of the roadway decking.
A crate is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal. The coefficient of kinetic friction between the crate and the ramp is 0.360. Find the acceleration of the moving crate.
Answer:[tex]1.95 m/s^2[/tex]
Explanation:
Given
inclination [tex]\theta =30.6^{\circ}[/tex]
coefficient of kinetic friction [tex]\mu =0.36 [/tex]
As crate is moving Down therefore friction will oppose the motion
using FBD
[tex]mg\sin \theta -f_r=ma [/tex]
[tex]f_r=\mu N[/tex]
[tex]f_r=\mu mg\cos \theta [/tex]
[tex]mg\sin \theta -\mu mg\cos \theta =ma[/tex]
[tex]a=g\sin \theta -\mu g\cos \theta [/tex]
[tex]a=g(\sin (30.6)-0.36\cdot \cos (30.6))[/tex]
[tex]a=9.8\times 0.199[/tex]
[tex]a=1.95 m/s^2[/tex]
To determine the crate's acceleration down the ramp, we calculate and subtract the kinetic friction from the component of the gravitational force acting down the slope. We then divide the resulting net force by the mass of the crate.
Explanation:To find the acceleration of the crate, we need to note that there are two forces acting on the crate as it moves down the ramp: the force due to gravity and the force due to friction. Since these two forces act in opposite directions, the crate's acceleration will be less than if there were no friction.
First, we need to calculate the component of the gravitational force acting down the slope, which can be found using the formula mg sin θ, where m is the mass of the crate, g is the acceleration due to gravity, and θ is the angle of the incline.
Next, we calculate the kinetic friction using the formula μk mg cos θ, where μk is the coefficient of kinetic friction.
The net force acting on the crate is the difference between these two forces and since F = ma, where F is the net force and m is the mass of the crate, we can solve for acceleration (a) by dividing the net force by the mass of the crate.
This calculation will give us the acceleration of the crate down the incline, taking into account the presence of friction.
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A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straight down the slope. If the slope were frictionless, the cylinder would slide down the slope without rotating. What minimum coefficient of static friction is needed for the cylinder to roll down without slipping?
Answer:
[tex]\mu_s=\frac{1}{3}\tan \theta[/tex]
Explanation:
Let the minimum coefficient of static friction be [tex]\mu_s[/tex].
Given:
Mass of the cylinder = [tex]M[/tex]
Radius of the cylinder = [tex]R[/tex]
Length of the cylinder = [tex]L[/tex]
Angle of inclination = [tex]\theta[/tex]
Initial velocity of the cylinder (Released from rest) = 0
Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.
We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,
[tex]\tau =I\alpha[/tex]
Now, angular acceleration is given as:
[tex]\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}[/tex]
Also, moment of inertia for a cylinder is given as:
[tex]I=\frac{MR^2}{2}[/tex]
Therefore, the torque acting on the cylinder can be rewritten as:
[tex]\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1[/tex]
Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are [tex]mg\sin \theta\ and\ f[/tex]. The net force acting along the incline is given as:
[tex]F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2[/tex]
Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.
So, [tex]N=Mg\cos \theta[/tex]
Plugging in [tex]N=Mg\cos \theta[/tex] in equation (2), we get
[tex]F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3[/tex]
Now, as per Newton's second law,
[tex]F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4[/tex]
Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.
[tex]\tau=fR\\\frac{MRa}{2}=\mu_sMg\cos \theta\times R\\\\a=2\times \mu_sg\cos \theta\\\\But, a=g(\sin \theta-\mu_s \cos \theta)\\\\\therefore g(\sin \theta-\mu_s \cos \theta)=2\times \mu_sg\cos \theta\\\\\sin \theta-\mu_s \cos \theta=2\mu_s\cos \theta\\\\\sin \theta=2\mu_s\cos \theta+\mu_s\cos \theta\\\\\sin \theta=3\mu_s \cos \theta\\\\\mu_s=\frac{\sin \theta}{3\cos \theta}\\\\\mu_s=\frac{1}{3}\tan \theta............(\because \frac{\sin \theta}{\cos \theta}=\tan \theta)[/tex]
Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:
[tex]\mu_s=\frac{1}{3}\tan \theta[/tex]
The minimum coefficient of static friction that needed for cylinder to roll down without slipping is [tex]\mu_s= \frac{tan\theta}{3}[/tex]
Explanation:A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straight down the slope. If the slope were frictionless, the cylinder would slide down the slope without rotating. What minimum coefficient of static friction is needed for the cylinder to roll down without slipping?
Given: radius R, length L, angle θ, and mass M
We need to calcuate the minimum static friction coefficient. It is useful so the cylinder will roll without slipping down the incline. The cylinder is also released from rest. As the cylinder is rolling, we have to consider the moment of inertia. Rolling of cylinder is happened due to the friction force
By applying Newton law of motion
[tex]F = M a\\\tau = I \alpha\\\tau = I \frac{a}{R} \\\tau = \frac{1}{2} M R^2 \frac{a}{R}[/tex]
From diagram
[tex]Mg sin\theta - f_{fr} = Ma\\f_{fr} = \mu_s N\\f_{fr} = \mu_s Mg cos \theta\\a = g sin \theta - \mu_s cos \theta[/tex]
Then also
[tex]\tau = f_{fr} R\\f_{fr} = \frac{Ma}{2} \\\mu_s Mg cos \theta = \frac{Mg (sin\theta - \mu_s cos \theta)}{2} \\\frac{3}{2} \mu_s cos\theta = \frac{sin\theta}{2}\\ \mu_s = \frac{tan\theta}{3}[/tex]
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