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Artemon [7]
3 years ago
14

Two children of about the same weight are playing at the playground. They both climb up to the top of a small tower. One slides

back down to ground level on a frictionless slide. The other freely drops down from the tower. Both children start their trips back down to the ground with an initial speed of zero.
Which combination of statements below is most accurate? [Enter your answer as a string of the letter options you believe are correct. For instance, if you think options A, D, and E are correct, then enter "ADE"]

A) The work done by gravity on each child is positive as they climb the tower.
B) The children have different kinetic energies on reaching ground level
C) The work done by gravity on both children is negative as they descend from the tower.
D) The overall work done by gravity is zero.
E) None of the above
Physics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Answer:

D

Explanation:

D) The overall work done by gravity is zero  

This statement is correct .

If m be the mass of each of the children and h be the height of tower

work done by gravity on the boys in going up = - mgh

it is so because force applied by gravity = mg downwards and displacement

is upwards

work done will be negative = - mgh

Work done by gravity on boys when they come down = + mgh because both force and displacement are downwards .

Hence total work done = - mgh + mgh = 0.

The children will have same kinetic energy as the inclined surface is friction-less so no energy will be dissipated hence addition of energy to boys in both the cases will be same.

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Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
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