I am going to say velocity because you have the same amount of acceleration going in a certain direction.
Answer:
17304 J
Explanation:
Complete statement of the question is :
In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .
Part A
How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?
Solution :
= tension force in the tow rope = 350 N
= length of the incline surface = 120 m
= work done by tension force = ?
The tension force acts parallel to incline surface, hence work done by tension force is given as

= height gained by the rider = 30 m
= total mass of rider and tube = 84 kg
Potential energy gained is given as

= Thermal energy created
Using conservation of energy

Answer:

Explanation:
Here two charges are placed at distance "d" apart
now the net value of electric field at some position between two charges will be ZERO
so we will have
electric field due to charge 1 = electric field due to charge 2

Let the position where net field is zero will lie at distance "r" from q1

now we will have

now square root both sides

now we have

so we have

Answer:
The balloon prohibits the flow of air through the air capacitor.
Explanation:
Just like an electric capacitor has an insulator between the plates, the air capacitor has a balloon between the chambers.
Answer:
When you jump off a train, you jump off a certain height and your downwards (vertical) velocity is zero. But your forward (horizontal) velocity is not. You will hit the ground on split second with your horizontal velocity practically the same as the train.
Explanation:
you be in serious injury.