PHYSICS COLLEGE

If R = 12 cm, M = 430 g, and m = 60 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)

Answers

Answer 1

Answer:

Answer:

Explanation:

Given

Radius of Pulley r=12 cm

mass of block m=60 gm

mass of Pulley M=430 gm

Block descend h=50 cm

Applying Conservation of Energy

Potential Energy of block convert to rotational Energy of pulley and kinetic energy of block

i.e.

$mgh=(1)/(2)I\omega ^2+(1)/(2)mv^2$

where I=moment of inertia

$I=mr^2$

and for rolling $\omega =(v)/(r)$

$mgh=(1)/(2)Mv^2+(1)/(2)mv^2$

$v^2=(2mgh)/(m+M)$

$v=\sqrt{(2mgh)/(m+M)}$

$v=\sqrt{(2* 60* 9.8* 0.5)/(430+60)}$

$v=\sqrt{(60* 9.8)/(490)}$

$v=√(1.2)$

$v=1.095 m/s$

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Why is wood a renewable resource?A) We use it over and over again.
B) It takes millions of years to replace.
C) Trees grow everywhere, so we always have some.
D) It can be replaced in a reasonable amount of time.

Answers

Answer:

D) It can be replaced in a reasonable amount of time.

Explanation:

i took the test!!!!!1

Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringeFor what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?
Express your answer in micrometers(not in nanometers).

Answers

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

$\lambda$ = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

$tan\theta=(y)/(D)\n\Rightarrow \theta=tan^(-1){(y)/(D)}\n\Rightarrow \theta=tan^(-1){(4.84* 10^(-3))/(3)}\n\Rightarrow \theta=0.09243\ ^(\circ)$

$sin\theta=(\lambda)/(d)\n\Rightarrow d=(\lambda)/(sin\theta)\n\Rightarrow d=(598* 10^(-9))/(sin0.09243)\n\Rightarrow d=0.00037066\ m$

For first dark fringe

$dsin\theta=(\lambda')/(2)\n\Rightarrow \lambda'=2dsin\theta\n\Rightarrow \lambda'=2* 0.00037066* sin0.09243\n\Rightarrow \lambda'=1.196* 10^(-6)\n\Rightarrow \lambda'=1.196\ \mu m$

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

Final answer:

The wavelength of light that will produce the first-order dark fringe at the same point on the screen is the same as the original wavelength of the light, which is 598 nm (0.598 μm).

Explanation:

To find the wavelength of light that will produce the first-order dark fringe at the same point on the screen, we can use the equation dsinθ = nλ, where d is the separation between the slits, θ is the angle of the fringe, n is the order of the fringe, and λ is the wavelength of the light.

In this case, the first-order bright fringe is located at a distance of 4.84 mm from the center of the central bright fringe. Since this is a first-order fringe, n = 1.

Plugging in the values, we have (0.120 mm)(sinθ) = (1)(λ). Rearranging the equation gives sinθ = λ/0.120 mm.

Since the first-order dark fringe is located at the same point as the first-order bright fringe, the angle of the first-order dark fringe can be calculated by taking the sine inverse of λ/0.120 mm.

Finally, to find the wavelength of light that will produce the first-order dark fringe at this point, we can rearrange the equation to solve for λ: λ = (0.120 mm)(sinθ).

Now, substitute the known values into the equation to calculate the wavelength of light:

λ = (0.120 mm)(sinθ) = (0.120 mm)(sin sin^-1(λ/0.120 mm)) = λ.

The wavelength of light that will produce the first-order dark fringe at this point on the screen is the same as the original wavelength of light, which is 598 nm. Converting this value to micrometers, we get 0.598 μm.

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When a box is placed on an inclined surface with no friction, it will:

Answers

Answer: With no friction, the box will accelerate down the ramp

Explanation:

It will gain speed down the ramp

A skater has rotational inertia 4.2 kg-m2 with his fists held to his chest and 5.7 kg?m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they�re essentially on his rotation axis, how fast will he be spinning? Express your answer using two significant figures. ?f=

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Answer: 38.5rad/s

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Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a wavelength of 3.5 cm?Express your answer in GHz and using two significant figures.
f = ________GHz
Microwave signals are beamed between two mountaintops 52 km apart. How long does it take a signal to travel from one mountaintop to the other?
Express your answer in ms and using two significant figures.
t = ________ms

Answers

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =(c)/(\lambda) =(3e8m/s)/(0.035m) = 8.57e9 Hz (2)$

⇒ f = 8.6 Ghz (with two significative figures)

Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

$v =(d)/(t) (3)$

where v= c= speed of light in vacuum = 3*10⁸ m/s

d= distance between mountaintops = 52 km = 52*10³ m

Solving for t, we get:

$t = (d)/(c) = (52e3m)/(3e8m/s) = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)$

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Which exerts more force, the Earth pulling on the moon or the moon pulling on the Earth? Explain.

Answers

Answer: the earth

Explanation: Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth. Over a very long time, the moon's rotations created fiction with the Earth's tugging back, until the moon's orbit and rotational locked with Earth.

and that's why the earth pulls the moon

Final answer:

The Earth pulling on the moon and the moon pulling on the Earth exert the same amount of force on each other due to Newton's third law of motion.

Explanation:

In terms of force, the Earth pulling on the Moon and the Moon pulling on the Earth exert the same amount of force on each other. This is because of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. So, while the Earth's gravitational force pulls the Moon towards it, the Moon's gravitational force also pulls the Earth towards it with an equal amount of force.

Newton's third law of motion states that for every action, there is an equal and opposite reaction. In the context of the gravitational interaction between the Earth and the Moon, the forces they exert on each other are equal in magnitude and opposite in direction.

The Earth pulls on the Moon with a gravitational force, and, according to Newton's third law, the Moon simultaneously pulls on the Earth with an equal gravitational force. These forces are sometimes referred to as "action and reaction pairs." The force that the Earth exerts on the Moon is often called the gravitational attraction of the Earth on the Moon, and vice versa.

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