Answer:
has units of distance
has units of distance over time
has units of distance over 
has units of distance over 
Explanation:
Since the expression for the distance is:
then:
has units of distance
has units of distance over time
has units of distance over
has units of distance over
because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.
Answer: Contact force
a. Applying break in a vehicle.
d. The speed of ball rolling on ground is reduced
Non contact force
b. A coconut falling from a coconut tree.
c. The planets revolving around the sun.
Explanation:
The contact force is the force which exerts when one object or entity comes in contact with other object or entity. For example, on application of break the vehicle stops, the force is applied on the breaks to stop the vehicle. The ball rolling on the ground the speed reduces so the application of force on the ground also reduces.
The non contact force is the force one object exerts on the other without coming in direct contact with the other object. The force exerted by one object on other due to gravity is a non contact force. The coconut falling on the ground and planets revolving around the sun are examples of non contact force due to gravity.
Local winds are driven by temperature differences in areas fairly close to each other. If water and land absorbed and released heat at the same rate, there wouldn't be any temperature differences and nothing to power local winds. See the related link for further information.
Answer:
636.4 J
Explanation:
The potential energy between one of the charges at the corner of the square and the fifth identical charge is U = kq²/r where q = charge = +50 × 10⁻⁶ C and r = distance from center of square. = √2 m (since the midpoint of the sides = 1 m, so the distance from the charge at the corner to the center is thus √(1² + 1²) = √2)
Since we have four charges, the additional potential energy to move the charge to the centre of the square is U' = 4U = 4kq²/r
U' = 4kq²/r
= 4 × 9 × 10⁹ Nm²/C² (+50 × 10⁻⁶ C)²/√2 m
= 900 Nm²/√2 m
= 636.4 J
