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Yuki888 [10]
3 years ago
8

Sledding down a hill, you are traveling at 10 m/s when you reach the bottom. You (inertia 80 kg ) then move across horizontal sn

ow toward a 400-kg boulder but jump off the sled (inertia 8.0 kg ) the instant before it hits the boulder. The boulder is sitting on very slick ice and moves freely when the sled hits it. The sled bounces back, moving at 6.0 m/s. At what speed does the boulder move after the sled hits it?
Physics
1 answer:
RSB [31]3 years ago
5 0

Answer:

V = 0.32 m /s

Explanation:

Momentum of the man on the sledge along with sledge

= (80+8) x 10 = 880 kg. m/s When the man jumps off the sledge , the velocity of the sledge will remain intact at 10 m/s

When the sledge hits the boulder and bounces back the momentum of the sledge - boulder system will remain conserved .

change in momentum of sledge = 8 x (10 +6 )

= 128  kg m/s

Change in the momentum of boulder

= 400 V -0

= 400V

400V = 128

V = 0.32 m /s

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Answer:

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Explanation:

Given that,

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t=\dfrac{d}{v}\\\\t=\dfrac{1609}{3\cdot10^{8}}\\\\=5.36\times 10^{-6}\ s

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A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liqui
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Answer:

n_l = 1.97

Explanation:

given data:

refractive index of lens 1.50

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\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

focal length of lens in liquid is

\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2}   \right ]

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rearrange fron_l

n_l = \frac{n_g f_l}{f_l+f(n_g-1)}

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Answer:

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