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pshichka [43]
3 years ago
10

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1m/s^2 . The acce

leration period lasts for time 7.00s until the fuel is exhausted. After that, the rocket is in free fall. Find the maximum height y_max reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 rm{m/s^2} .
Physics
1 answer:
ollegr [7]3 years ago
7 0
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0 
y=44.1*100/2 = 2205m 
hence, the speed will be 
v=0 + a*t = 441m/s 
from that height it will just be subjected to the gravitational acceleration 
0=v_acc^2 -2g*y_free 
y_free = v_acc^2/2g = 9922.5m 
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
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A 13.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay s
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Answer:

v_{ic}=92.53 m/s

Explanation:

We need to apply conservation of momentum and energy to solve this problem.

<u>Conservation of momentum</u>

p_{i}=p_{f}

m_{c}v_{ic}=(m_{c}+m_{w})V (1)

  • m(c) is the mass of stick clay
  • m(w) is the mass of the wooden block
  • v(ic) is the initial velocity of clay
  • V is the final velocity of the system clay plus wood.

<u>Conservation of total energy</u>

The change in kinetic energy is equal to the change in internal energy, in our case it would be the energy loss due to the friction force. Let's recall the definition of work, it is the dot product between force and displacement, Therefore:

\Delta E=W

\frac{1}{2}(m_{c}+m_{w})V^{2}=F_{friction}*d

\frac{1}{2}(m_{c}+m_{w})V^{2}=\mu (m_{c}+m_{w})gd

We can find V from this equation:

V=\sqrt{2\mu gd}=\sqrt{2*0.65*9.81*7.5}=9.78 m/s

Now, let's put V into the equation (1) and find v(ic)

v_{ic}=\frac{(m_{c}+m_{w})V}{m_{c}}=\frac{123*9.78}{13}=92.53 m/s

I hope it helps you!  

<u />

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