All categories

3 years ago

A wall, acted upon by a force of 20 N, does not move. The work done on the wall in this process is

Physics

1 answer:

prisoha [69]3 years ago

5 0

Explanation:

A wall that is acted upon by a force of 20N does not move suggests that no work has been done on the wall.

Work done is the force applied to move in the direction of the applied force.

Work done = force x distance.

Since the distance is 0, the work done on the body too is zero.

Work is only done when forces moves a body through a given distance.

Learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

You might be interested in

Two samples of water are mixed together.

Anon25 [30]

<h3><u>Answer;</u></h3>

A.75°C

<h3><u>Explanation</u>;</h3>

Let the change in temp of cold water be x degrees,

while that of hot water be 100 - x degrees.

Heat exchange = mcΔt

Ice

Δt = x

m = 0.50 kg

c = 4.18 kJ/kg*°C

Hot water

Δt = 100 - x

m = 1.5 kg

c = 4.18

But;

Heat lost = heat gained

0.50 * c * x = 1.5 * c * (100 - x)

0.50 *x = 1.5*(100 - x)

0.5x = 150 - 1.5x

0.5x + 1.5x = 150 - 1.5x + 1.5x

2x = 150

x = <u>75° C</u>

Hence; the equilbrium temperature will be 75° C

7 0

3 years ago

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before

Tasya [4]

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, $d=2.5\times 10^{-15}\ m$

Speed of light, $c=3\times 10^{8}\ m\s$

Let t is the time interval required for the strong interaction to occur. The speed is given by :

$c=\dfrac{d}{t}$

$t=\dfrac{d}{c}$

$t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}$

$t=8.33\times 10^{-24}\ s$

So, the time interval required for the strong interaction to occur is $8.33\times 10^{-24}\ s$ . Hence, this is the required solution.

8 0

3 years ago

The layer of leaves that blocks most of the sunlight from reaching the ground in the rain forest is called the _____.

love history [14]

The answer should be <span>canopy.</span>

4 0

3 years ago

Read 2 more answers

If a 93000 kg truck collides with a 60 kg car

algol [13]

What’s the question here?

8 0

3 years ago

Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

6 0

2 years ago