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aniked [119]
3 years ago
14

For problems that involve an object accelerating along an inclined plane, how can the weight be used to determine the force comp

onent that causes motion? The weight is the full force that acts perpendicular to the surface of the ramp. The weight times the cosine of the ramp's angle is the force perpendicular to the ramp. The weight times the sine of the angle of the ramp acts down along the ramp. O The weight times the cosine of the angle of the ramp acts down along the ramp.​
Physics
1 answer:
irakobra [83]3 years ago
3 0

Answer:

C

Explanation:

Ed2020

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A sprinter accelerates from rest to 10.0 m/s in 1.28 s . Part A Part complete What is her acceleration in m/s2? a a = 7.81 m/s2
Mashutka [201]

Explanation:

It is given that,

Initial speed of sprinter, u = 0

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a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :

a=\dfrac{v-u}{t}

a=\dfrac{10\ m/s}{1.28\ s}

a=7.81\ m/s^2

b. Final speed of the sprinter, v = 36 km/h

Time, t = 0.000355 h

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statuscvo [17]

Answer:

The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

Given;

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K.E = ¹/₂mv²

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K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} =  (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} =  (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s

Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s

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