Answer:
The magnitude of the acceleration of the car is 35.53 m/s²
Explanation:
Given;
acceleration of the truck, 
= 12.7 m/s²
mass of the truck, 
= 2490 kg
mass of the car, 
= 890 kg
let the acceleration of the car at the moment they collided = 
Apply Newton's third law of motion;
Magnitude of force exerted by the truck = Magnitude of force exerted by the car.
The force exerted by the car occurs in the opposite direction.
Therefore, the magnitude of the acceleration of the car is 35.53 m/s²
Answer: 0.98m
Explanation:
P = -74 mm Hg = 9605 Pa = 9709N/m^2
= 9605 kg m/s^2/m^2
density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3
Pressure equation: P = rho g h
h = P/(rho g)
h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)
h = 0.98 m
0.98m is the maximum depth he could have been.
Answer: TRUST ME I GOT IT WRONG the answer is B
Explanation:
Answer:
A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.
Explanation:
This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂
Since F₁ = 20 N, then F₂ = -F₁ = -20 N
So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.
So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.
