The reacting force that is equal to and opposite in the direction to the centripetal force and tends to fling air out of the cen
1 answer:
You might be interested in
Answer:
The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s
Explanation:
The momentum of the particle is related to force by the following equation:
Δp = F · Δt
Where:
Δp = change in momentum = final momentum - initial momentum
F = constant force.
Δt = time interval.
Let´s calculate the x-component of the momentum after the 0.13 s:
final momentum - 8 kg m/s = -7 N · 0.13 s
final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s
final momentum = 7.09 kg m/s
Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:
final momentum = 5 kg m/s² · 0.13 s
final momentum = 0.65 kg m/s
Then, the mometum vector will be as follows:
p = (7.09 kg m/s, 0.65 kg m/s)
The magnitude of this vector is calculated as follows:
The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s
Explanation:
since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.
We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.
range can be calculated by the formula :-
u is the velocity during its take off and is the angle at which its thrown
Given that
- u = 8m/ s
= 40°
calculating range using the above formula
value of sin 80 = 0. 985
Hence,
Answer:
F= 600 N
Explanation:
Given that
Initial velocity ,u= 0 m/s
Final velocity ,v= 30 m/s
mass ,m = 0.5 kg
time ,t= 0.025 s
The change in the linear momentum is given as
ΔP= m (v - u)
ΔP= 0.5 ( 30 - 0 ) kg.m/s
ΔP= 15 kg.m/s
We know that from second law of Newtons
Now by putting the values
F= 600 N