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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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azamat

Answer:

Answer D : about 1067 meters

Explanation:

There are two steps to this problem:

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a=\frac{Vf-Vi}{t}

Where Vf is the final velocity of the plane (in our case: zero )

Vi is the initial velocity of the plane (in our case: 80 m/s)

a is the acceleration (in our case -3 m/s^2 - notice negative value because the velocity is decreasing)

a=\frac{Vf-Vi}{t}\\-3=\frac{0-80}{t}\\t=\frac{-80}{-3} = \frac{80}{3}

with units corresponding to seconds given the quantities involved in the calculation.

2) Second knowing the time it took the plane to stop, now use that time in the equation for the distance traveled under accelerated motion:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi= 80 (\frac{80}{3}) +\frac{1}{2} (-3) (\frac{80}{3}) ^{2}=1066.666666...

Where the answer results in units of meters given the quantities used in the calculation.

We round this to 1067 meters

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You knew that this question is ridiculously easy.  So, just to
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Read 2 more answers
an op amp in unity gain configuration (buffer) with slew rate of 5v/us is used to amplify a sinusoidal signal with a frequency o
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Answer:

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Explanation:

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Therefore, from the above equation we can write,

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