First of all, we can find the mass of the person, since we know his weight W:

And so

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:

There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as

We know the acceleration of the system,

(upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv
<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
Answer:
-10.9 rad/s²
Explanation:
ω² = ω₀² + 2α(θ - θ₀)
Given:
ω = 13.5 rad/s
ω₀ = 22.0 rad/s
θ - θ₀ = 13.8 rad
(13.5)² = (22.0)² + 2α (13.8)
α = -10.9 rad/s²
Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved
so we will have

now we know that




now we will have



so gun will recoil with speed 6.67 m/s