Answer:
The velocity when the ball hits the ground is obtained using v2. 2 = v1. 2 + 2 g Dy with v1=0 and Dy=h. Thus solving for v2 yields 17.1 m/s v2 = 2 g h =.
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Gay-Lussac's Law states
P1 / T1 = P2 / T2
So the answer is b
Answer:
q=3.5*10^-4
Explanation:
<u>concept:</u>
The force acting on both charges is given by the coulomb law:
F=kq1q2/r^2
the centripetal force is given by:
Fc=mv^2/r
The kinetic energy is given by:
KE=1/2mv^2
<u>The tension force:</u>
<u><em>when the plane is uncharged </em></u>
T=mv^2/r
T=2(K.E)/r
T=2(50 J)/r
T=100/r
<u><em>when the plane is charged </em></u>
T+k*|q|^2/r^2=2(K.E)charged/r
100/r+k*|q|^2/r^2=2(53.5 J)/r
q=√(2r[53.5 J-50 J]/k) √= square root on whole
q=√2(2)(53.5 J-50 J)/8.99*10^9
q=3.5*10^-4
Answer:
The value is 
Explanation:
From the question we are told that
The speed of the rope with hook is 
The angle is 
The speed at which it hits top of the wall is 
Generally from kinematic equation we have that
Here h is the height of the wall so
![[16.3 sin (65)]^2 = [24.1 sin (65)] ^2+ 2 (-9.8)* h](https://tex.z-dn.net/?f=%5B16.3%20sin%20%2865%29%5D%5E2%20%3D%20%20%5B24.1%20sin%20%2865%29%5D%20%5E2%2B%20%20%202%20%28-9.8%29%2A%20h)
=>
Answer:
C. It is radiation leftover from the Big Bang
