Gravity is affected by mass and distance
Answer:
(A) Distance will be equal to 1.75 km
(B) Displacement will be equal to 1.114 km
Explanation:
We have given circumference of the circular track = 3.5 km
Circumference is given by 
r = 0.557 km
(a) It is given that car travels from southernmost point to the northernmost point.
For this car have to travel the distance equal to semi perimeter of the circular track
So distance will be equal to 
(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track
So displacement will be equal to d = 2×0.557 = 1.114 m
Answer:
Resistance is inversely proportional to current, so when the resistance doubles, the current is cut in half. Resistance is directly proportional to current, so when the resistance doubles, the current is cut in half.
Explanation:
It is given that,
Mass of Millersburg Ferry, m = 13000 kg
Velocity, v = 11 m/s
Applied force, F = 10⁶ N
Time period, t = 20 seconds
(a) Impulse is given by the product of force and time taken i.e.



(b) Impulse is also given by the change in momentum i.e.





(c) For new velocity,



Hence, this is the required solution.
Assume the motion when you are in the car or in the school bus to go to the school.
To describe the motion the first thing you need is a point of reference. Assume this is your house.
This should be a description:
- When you are sitting and the car has not started to move you are at rest.
- The car starts moving from rest, gaining speed, accelerating. You start to move away from your house, with a positive velocity (from you house to your school) and positive acceleration (velocity increases).
- The car reaches a limit speed of 40mph, and then moves at constant speed. The motion is uniform, the velocity is constant, positive, since you move in the same direction), and the acceleration is zero.
- When the car approaches the school, the driver starts to slow down. Then, you speed is lower but yet the velocity is positive, as you are going in the same direction. The acceleration is negative because it is in the opposite direction of the motion.
- When the car stops, you are again at rest: zero velocity and zero acceleration.
- In all the path your velocity was positive, constant at times (zero acceleration) and variable at others (accelerating or decelerating).
- When you comeback home, then you can start to compute negative velocities, as you will be decreasing the distance from your point of reference (your house).