Examples of Newton's 2nd Law If you use the same force to push a truck and push a car, the car will have more acceleration than the truck, because the car has less mass. It is easier to push an empty shopping cart than a full one, because the full shopping cart has more mass than the empty one.
Answer:
The correct option is momentum is conserved
Explanation:
A closed system is a system that is independent/free of external factors/force and does not exchange matter with its surrounding. Since a close system is free of external factors/force; <em>acceleration is constant in it, mass is conserved in it and there will be changes in velocities of objects in the closed system</em>.
This question actually seeks to test our knowledge of the law of momentum. The law of conservation of momentum states that the momentum of a closed system is conserved.
Answer:
If the cart is being pushed at a constant speed, then the acceleration in the direction of motion is zero. Hence, the force in the direction of the motion is zero, according to Newton's Second Law.
For simplicity, I will denote the direction along the inclined ramp as x-direction.
In the question the value of the force is not clearly given, so I will denote it as 
Here the angle between the applied force and the x-direction is 12° + 17° = 29°
The x-component of the weight of the cart is equal to sine component of the weight.
Since the cart is rolling on tires the kinetic friction does no work.
Work done by the applied force:
Work done by the weight of the cart:
Since the x-component of the weight is in the -x-direction, its work is negative.
Conveniently, the total work done on the particle is zero, since its velocity is constant.
That reactants are favored.
Answer:
the runner's average kinetic energy during the run is 476.96 J.
Explanation:
Given;
mass of the runner, m = 85 kg
distance covered by the runner, d = 42.2 km = 42,200 m
time to complete the race, t = 3 hours 30 mins = (3 x 3600s) + (30 x 60s)
= 12,600 s
The speed of the runner, v = d/t
v = 42,200 / 12,600
v = 3.35 m/s
The runner's average kinetic energy during the run is calculated as;
K.E = ¹/₂mv²
K.E = ¹/₂ × 85 × (3.35)²
K.E = 476.96 J
Therefore, the runner's average kinetic energy during the run is 476.96 J.
