haavhiinsgys sujan usnugs guisjg suhab rojina
A ) The first ball:
h = v o * t - g t² / 2
19.6 = 14.7 t + 9.8 t²/2
4.9 t² + 14.7 t - 19.6 = 0
t 1/2 = (-14.7 +/- √216.09+384.16 ) / 9.8 = 9.8/9.8 = 1 s
The second ball:
v = v o - g t
0 = 14.7 - 9.8 t
9.8 t = 14.7
t = 1.5 ( upward )
h = 19.6 + 14.7 - 9.8 * 1.5 = 30.625 m
30.625 = 9.8 * t²/2
t² = 6.25
t = 2.5 s ( downward )
1.5 s + 2.5 s = 4 s
The difference: 4 s - 1 s = 3 s.
B ) The first ball:
v = v o + g t = 14.7 + 9.8 * 1 = 24.5 m/s
The second ball:
v = g * t = 9.8 * 3 = 24.5 m/s
C ) d 1 = 14.7 * 0.8 + 9.8 * 0.8²/2 = 11.76 + 3.136 = 14.896 m
d 2 = 14.7 * 0.8 - 9.8* 0.8²/2 = 11.76 - 3.136 = 8.624 m
d 1 + d 2 = 14.896 + 8.624 = 23.52 m
Answer:
Solved
Explanation:
Free body diagram has been attached
and FBD we can say that
tension in c and b will be higher as T_1 helps to keep both the person steady.
So,
m_a=m_b=m_c= 65 kg and θ=25°
The term g×sinθ is the component of acceleration along the inclined surface
Similarly
putting the value of θ and mass we get
T_2= 269.207 N
T_1= 538.41 N
Answer:work done can be calculated by the following formula;
Explanation:
Answer:
8.2 m/s
Explanation:
radius (r) = 1.3 m
coefficient of friction (u) = 0.19
acceleration due to gravity (g) = 9.8 m/s^{2}
for the passengers to remain stuck on the wall, the normal force must be equal to the centripetal force (while considering the coefficient of friction).
normal force = centripetal force x coefficient of friction
mg = x u
g = x u
v =
v =
v = 8.2 m/s
the minimum speed the passengers can have to stick to the wall = 8.2 m/s