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2 years ago

10

A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A is at the top of the circle; C

is at the bottom. Points B and D are exactly halfway between A and C. Which one of the following statements concerning the A boy is whirling a stone at the end of a string around his head. The string makes one complete revolution every second, and the tension in the string is FT. The boy increases the speed of the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions per second. What happens to the tension in the sting?

1 answer:

mrs_skeptik [129]2 years ago

3 0

Answer:

Tension in the string will increase

Explanation:

As we know that tension in the string at any angle with the vertical is given as

$T - mgcos\theta = m\omega^2 R$

now we have

$T = mgcos\theta + m\omega^2 R$

also we know that

angular speed of the stone is directly depending on the time period of the motion

so it is given as

$\omega = \frac{2\pi}{T}$

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s

so the angular speed would be doubled

So here we can say that

tension in the string will increase when we will increase the frequency of revolution.

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The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a

SashulF [63]

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm= $1900\times 10^{-9}m$

Where $1nm=10^{-9} m$

$\lambda_2=400nm=400\times10^{-9}m$

$\lambda_1=700nm=700\times 10^{-9}m$

We have to find the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

$\theta=sin^{-1}(\frac{m\lambda}{d})$

Using the formula

m=1

$\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})$

$\theta=21.62^{\circ}$

Now, m=2

$\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})$

$\theta_2=24.90^{\circ}$

$\Delta \theta=\theta_2-\theta_1$

$\Delta \theta=24.90-21.62$

$\Delta \theta=3.28^{\circ}$

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

6 0

2 years ago

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

$v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft$

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0

3 years ago

As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a

Annette [7]

Im pretty sure it’s a because it makes more sense you know?.

5 0

2 years ago

Most of our Solar system is composed of___

Bond [772]

Answer:

empty space

Explanation:

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Hence, it can be concluded that Most of our Solar system is composed of "Empty Spaces."

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2 years ago

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Phantasy [73]

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3 years ago

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