Answer:
I will choose the answer A
Answer:
750 J
Explanation:
We have a student that pushes a 50N block across the floor for a distance of 15m. The question is asking how much work was done to move the block.
To solve this, we must know that we are looking for a certain thing called joules. And to get the answer, we must follow the formula of W = FS
F being the force and S being the distance.
W = FS
W = (50)(15)
W = 750
Therefore, 750 joules is our answer.
Answer:
A. 4.47 m/s
Explanation:
As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m
1500 g = 1.5 kg
Answer:
The answer is based on the conservation of energy law; something you should really understand by now.
For convenience we can hold one of the two charges still; it becomes the frame of reference. And everything we say is in reference to the designated static charge, call it Q.
So the moving charge, call it q, has total energy TE = PE. It's all potential energy as we start with q not moving.
It has potential energy because in order to separate q from Q, we had to do work, add energy, on q. And from the COE law, that work added is converted into PE.
It's a bit like lifting something off the ground. That's work and it becomes GPE. So there's some work, in separating the two charges in the first place.
But there's more.
Now we let q go. As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.
And there's more work, done by the EMF on charge q. That converts the PE into KE and the q charge smashes into Q with some kinetic energy.
Answer:
Explanation:
let the ladder is of mass "m" and standing at an angle with the ground
So here by horizontal force balance we will have
by vertical force balance we have
now by torque balance about contact point on ground we will have
so we will have
now from first equation we have
