A. 314 because when you use the formula for the GPE ; GPE=MGH or means mass times gravity time height (4x8x9.8) and thats equivalent to 313.6 which rounds up to 314. Hope it helps

8 0

4 years ago

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A ball rolls from rest at the top of a 15.5 m hill, downward to the right and back upward onto a 8.25 m hill. How fast is the ba

lana66690 [7]

Answer:

Explanation:

6 0

3 years ago

An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place

anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

$mu=(M+m)v$

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e. $\frac{v_1}{v_2}=1.084$

4 0

3 years ago