Use the formula for distance dropped from rest:
D = 1/2 a t^2 .
With no air resistance, you get about 8.8 sec.
I'm pretty sure you can fill in the missing work.
Answer:
Explanation:
Given that
At 17°C ,Lo= 20.09 cm
At 306°C ,L= 20.18 cm
ΔL= 200.18 - 20.09 = 0.09 cm
ΔT = 306 - 17 = 289°C
We know that
ΔL = L α ΔT
α =coefficient of linear expansion
ΔL = Lo x α x ΔT
0.09 = 20.09 x α x 289
Coefficient of linear expansion,
Answer:
The velocity of ball and man after catching ball is
Explanation:
Since there is no friction and no external force the momentum is conserved .
Here when man catches ball then man and ball move with common velocity .
Let the final common velocity be ,
Given that mass of the ball is =
Given that mass of man is
Initial velocity of ball =
Now considering momentum conservation
I believe the answer is potential energy if i remember correctly.
Answer:
a) 16.32 m/s
b) 640 N
Explanation:
A) mass of rocket m_r = 1000 g = 1 kg
initial speed of rocket u_r = 15 m/sec
initial speed of ball is u_b = 18 m/sec
final speed of ball is v_b = 40 m/sec
Let m_b be the mass of the ball= 60 g and v_r be the final velocity of the rocket
from law of conservation of momentum
momentum of the system remains zero
m_r×(u_r-v_r)+m_b(16-42) = 0
1×(15-v_r) = -0.060(18-40)
15-v_r = -1.32
v_r = 15+1.32 = 16.32 m/sec.
B) Average force that the rocket exert's on the ball is F_avg can be calculated as
contact time t=7.00 ms
F_avg = m(v-u)/t = 0.06×(40+18)/0.007 = 640 N