Answer:
Explanation:
Initial speed, v = 10 x 10^3 m/s
Mass of the earth, M = 6 x 10^24 kg
Radius of the earth, R = 6.4 x 10^6 m
Maximum from the surface of earth, h = ?
Let m = Mass of the projectile
Solution:
Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface
=
=
Answer:
58.5 m
Explanation:
First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.
The initial vertical velocity of the ball is
where
u = 21.5 m/s is the initial speed
is the angle
Substituting,
The vertical position of the ball at time t is given by
where
h = 13.5 m is the initial heigth
is the acceleration of gravity (negative sign because it points downward)
The ball reaches the water when y = 0, so
Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.
The horizontal velocity of the ball is
And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:
The first opiton is the answer A)<span>Rahul’s weight
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Answer: Choose the normal force acting between the object and the ground. Let's assume a normal force of 250 N.
Determine the friction coefficient.
Multiply these values by each other: 250 N * 0.13 = 32.5 N .
You just found the force of friction!
Explanation:
Question is from B to C
Answer: (b) 1.5m/s
x1=3m, x2=9m
t1=1s, t2=5s
Displacement, ∆x=(9-3)m=6m
Time elapsed, ∆t=(5-1)s=4s
So average velocity v =∆x/∆t=6/4=1.5m/s