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fredd [130]
3 years ago
14

In midair in the international space station a 1 kg chunk of putty moving at 1 m/s collides with and sticks to a 5 kg chunk of p

utty initially at rest. at what speed does the 6 kg chunk travel after the collision?
Physics
1 answer:
Delicious77 [7]3 years ago
6 0
<span>We know that the momentum keeps constant in a inelastic collisions, so the product of mass and speed do not change:
   m1 * v1 + m2 * v2 = m * v
 1 * 1 + 5 * 0 = (1 + 5) * v
  1 = 6 * v
 v = 1/6 m/s
   So the final speed of the 6 kg chunk will travel at 0.167 m/s</span>
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A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha
lord [1]

Answer:

The value is F_f =  46.935 \  N

Explanation:

From the question we are told that

    The  magnitude of the horizontal force is F  =  92.7 \  N

     The mass of the crate is  m  =  40.5 \  kg

     The acceleration of the crate is  a =  1.13 \ m/s

Generally the net force acting on the crate is mathematically represented as

       F_{net} =  F -  F_f =  ma

Here F_f is force of kinetic friction (in N) acting on the crate

      So  

            92.7  -  F_f =  40.5 * 1.13

=>         F_f =  46.935 \  N

5 0
3 years ago
A gymnast is swinging on a high bar. The distance between his waist and the bar is 0.905 m, as the drawing shows. At the top of
Citrus2011 [14]

Answer:

5.959 m/s

Explanation:

m = Mass of gymnast

u = Initial velocity

v = Final velocity

h_i = Initial height

h_f = Final height

From conservation of Energy

\frac{1}{2}mv^2+mgh_f=\frac{1}{2}mu^2+mgh_i\\\Rightarrow\frac{1}{2}mv^2+mg0=\frac{1}{2}m0^2+mgh_i\\\Rightarrow \frac{1}{2}mv^2=mgh_i\\\Rightarrow v=\sqrt{2gh_i}

h_i=2r

v=\sqrt{4gr}\\\Rightarrow v=\sqrt{4\times 9.81\times 0.905}\\\Rightarrow v=5.959\ m/s

Velocity of gymnast at bottom of swing is 5.959 m/s

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4 years ago
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Answer:

the answer is

B)It was the first planet located through mathematical calculations

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