Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?

Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.

Therefore, the ball will reach 20 cm before it starts to roll back down.

7 0

3 years ago

In a darkened room, a burning candle is placed 1.84 m from white wall. A lens is placed between candle and wall at a location th

KATRIN_1 [288]

Answer:

82.4 cm

Explanation:

The object and screen are kept fixed ie the distance between them is fixed and by displacing lens between them images are formed on the screen . In the first case let u be the object distance and v be the image distance

then ,

u + v = 184 cm

In the second case of image formation , v becomes u and u becomes v only then image formation in the second case is possible.

The difference between two object distance ie( v - u ) is the distance by which lens is moved so

v - u = 82.4 cm

7 0

3 years ago

How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners

Inessa05 [86]

Answer:

1.38 x 10^-18 J

Explanation:

q = - 1.6 x 10^-19 C

d = 5 x 10^-10 m

the potential energy of the system gives the value of work done

The formula for the potential energy is given by

$U =\frac{Kq_{1}q_{2}}{d}$

So, the total potential energy of teh system is

$U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}$

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

$U =3\times \frac{Kq^{2}}{d}$

K = 9 x 10^9 Nm^2/C^2

By substituting the values

$U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}$

U = 1.38 x 10^-18 J

6 0

3 years ago