Suppose two comets, comet A and comet B, were orbiting the Sun, having the same average orbital radii. If comet A had a higher e
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Answer:
Force that acted on the body was F = 13 N
Explanation:
If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s
When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:
which in our case becomes;
and we can solve for the acceleration as:
a = 10/10 m/s^2 = 1 m/s^2
Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:
F = 13 kg * 1 m/s^2 = 13 N
Answer:
Speed of the ball relative to the boys: 25 km/h
Speed of the ball relative to a stationary observer: 35 km/h
Explanation:
The RV is travelling at a velocity of
Here we have taken the direction of motion of the RV as positive direction.
The boy sitting near the driver throws the ball back with speed of 25 km/h, so the velocity of the ball in the reference frame of the RV is
with negative sign since it is travelling in the opposite direction relative to the RV. Therefore, this is the velocity measured by every observer in the reference frame of the RV: so the speed measured by the boys is
v = 25 km/h
Instead, a stationary observer outside the RV measures a velocity of the ball given by the algebraic sum of the two velocities:
v = +60 km/h + (-25 km/h) = +35 km/h
So, he/she measures a speed of 35 km/h.
To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.
Here,
v = Final velocity
= Initial velocity
g = Acceleration due to gravity
t = Time
At t = 4s, v = -30m/s (Downward)
Therefore the initial velocity will be
Now the position can be calculated as,
When it has the ground, y=0 and the time is t=4s,
Therefore the cliff was initially to 41.6m from the ground