Answer:
As we need to use a nested loop in our function,hence push $ra
pop $ra
jal nested_function_label
nop is the correct option.
A dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s due to acceleration due to gravity.
s = vt - 1 / 2 at²
s = Displacement
v = Final velocity
t = Time
a = Acceleration
s = 5 m
t = 1 s
a = 10 m / s²
5 = ( v * 1 ) - ( 1 / 2 * 10 * 1 * 1 )
5 = v - 5
v = 10 m / s
The equation used to solve the given problem is an equation of motion. In a free fall motion, usually air resistance is not considered for easier calculation. If air resistance is considered acceleration cannot be constant throughout the entire motion.
Therefore, a dropped object only fall 5 meters down after 1 second of freefall, yet achieve a speed of 10m/s due to acceleration due to gravity.
To know more about equation of motion
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F = ma
We have mass = 0.2kg
and acceleration = 20 m/s^2
So..
F = (0.2)(20)
F = 4 N
Answer:
(a) v = 3..6 m/s
(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.
Explanation:
from the question we have the following:
mass of the car (Mc) = 24,000 kg
initial velocity of the car (u) = 4 m/s
mass of water (Mw) = 3000 kg
final velocity of the car (v) = ?
(a) we can calculate the final momentum of the car by applying the conservation of momentum where
initial momentum = final momentum
Mc x U = (Mc + Mw) x V
24000 x 4 = (24000 + 3000) x v
96,000 = 27000v
v =3.6 m/s
(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.
