I would say your answer is B, since Newton's 3rd law is, "For every action, there is an equal and opposite reaction."
It's talking about pairs of actions. Sorry if I'm wrong.
The force required to slow the truck was -5020 N
Explanation:
First of all, we find the acceleration of the truck, which is given by

where
v is the final velocity
u is the initial velocity
t is the time
For the truck in this problem,
v = 11.5 m/s
u = 21.9 m/s
t = 2.88 s
So the acceleration is

where the negative sign means that this is a deceleration.
Now we can find the force exerted on the truck, which is given by Newton's second law:

where
m = 1390 kg is the mass of the truck
is the acceleration
And substituting,

So the closest answer among the option is -5020 N.
Learn more about acceleration and forces:
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Answer:
m=417.24 kg
Explanation:
Given Data
Initial mass of rocket M = 3600 Kg
Initial velocity of rocket vi = 2900 m/s
velocity of gas vg = 4300 m/s
Θ = 11° angle in degrees
To find
m = mass of gas
Solution
Let m = mass of gas
first to find Initial speed with angle given
So
Vi=vi×tanΘ...............tan angle
Vi= 2900m/s × tan (11°)
Vi=563.7 m/s
Now to find mass
m = (M ×vi ×tanΘ)/( vg + vi tanΘ)
put the values as we have already solve vi ×tanΘ
m = (3600 kg ×563.7m/s)/(4300 m/s + 563.7 m/s)
m=417.24 kg
Explanation:
It is given that,
Mass of the tackler, m₁ = 120 kg
Velocity of tackler, u₁ = 3 m/s
Mass, m₂ = 91 kg
Velocity, u₂ = -7.5 m/s
We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,


v = -1.5 m/s
Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.
Answer:
Explanation:
We know that the pressure can be calculated in the following way:
p = d·g·h
with d being the density of the water, g the gravitational acceleration and h the depth.
Also d of the water = 1000 kg/m^3 circa and g = 9.8 m/s^2 circa
117,500 Pa = 1000kg/m³ · 9.8m/s² · h
Therefore h = 11,9 m