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Genrish500 [490]
3 years ago
6

B) Write the name of the deviceswhich are shown in the givenfigures and write their uses​

Physics
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

transparent Prism

Explanation:

The device shown is a prism. It is used to study the refraction of light according with its wavelength as it transverses from a medium (air for example) to another (either glass, fluorite, plastic), and vice versa.

For that reason is also used to separate white light into its spectral wavelength components.

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Red - sensitive so answer is c
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On Mars a rock falls an unknown vertical distance from a resting position and lands in a crater. If it takes the rock 2.5 second
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The Answer To This Question Is B

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A skateboarder starting from rest accelerates down a ramp at 2 m/s for 2 s. What is the final speed of the skateboarder?
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3 years ago
(II) You buy a 75-W lightbulb in Europe, where electricity is delivered at 240 V. If you use the bulb in the United States at 12
Elodia [21]

Answer:

Explanation:

You are looking for the resistance to start with

W = E * E/R

75 = 240 * 240 / R

75 * R = 240 * 240

R = 240 * 240 / 75

R = 57600 / 75

R = 768

Now let's see what happens when you try putting this into 110

W = E^2 / R

W = 120^2 / 768

W = 18.75

So the wattage is rated at 75. 18.75 is a far cry from that. I think they intend you to set up a ratio of

18.75 / 75 = 0.25

This is the long sure way of solving it. The quick way is to realize that the voltage is the only thing that is going to change. 120 * 120 / (240 * 240) = 1/2*1/2 = 1/4 = 0.25

4 0
3 years ago
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
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