Answer:
when it hit the moving bat
Explanation:
force equals mass times acceleration which means the moving bat will add more force to the ball.
<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal
it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
--------------------------------------...
let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
=====================
t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s
so it will able to cross</span>
Answer:
31.42383 m/s
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
= Coefficient of kinetic friction = 0.48
s = Displacement = 0.935 m
= Mass of bean bag = 0.354 kg
= Mass of empty crate = 3.77 kg
= Speed of the bean bag
= Speed of the crate
Acceleration
From equation of motion
In this system the momentum is conserved
The speed of the bean bag is 31.42383 m/s
Three 40w lamp for 6 hours
Solution :
Let 
kg
m/s
Let 
and 
are the speeds of the disk 
and 
after the collision.
So applying conservation of momentum in the y-direction,
Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.
Now applying conservation of momentum in the x-direction,
m/s
So, 
= 4.33 m/s
Therefore, speed of the disk 2 after collision is 4.33 m/s
