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lubasha [3.4K]
3 years ago
11

Can someone please help me? If somebody pushes a cart with 4N of force and the cart moves a distance of 2 meters, how much work

did they do?
Physics
1 answer:
sergeinik [125]3 years ago
7 0

Answer:

Work done, W = 8 J

Explanation:

We have,

Force acting on the cart is 4 N

Cart covers a distance of 2 m

We need to find the work done by the cart. Work done by an object is given in terms of force and displacement. It can be given by :

W=Fd\\\\W=4\ N\times 2\ m\\\\W=8\ J

So, the work done by the cart is 8 J.

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What force does it take to accelerate a 9.2 kg object 7.0 m/s^2?
Ulleksa [173]

Answer:

<h2>64.4 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 9.2 kg

acceleration = 7 m/s²

We have

force = 9.2 × 7 = 64.4

We have the final answer as

<h3>64.4 N</h3>

Hope this helps you

5 0
3 years ago
HELP MY GRADES-
belka [17]
If you count the number of seconds between the flash of lightning and the sound of thunder, and then divide by 5, you'll get the distance in miles to the lightning: 5 seconds = 1 mile, 15 seconds = 3 miles, 0 seconds = very close. Keep in mind that you should be in a safe place while counting.
5 0
3 years ago
A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away
alekssr [168]

Answer:

31.2 m/s

Explanation:

f_{app} = Frequency of approach = 480 Hz

f_{aw} = Frequency of going away = 400 Hz

V = Speed of sound in air = 343 m/s

v = Speed of truck

Frequency of approach is given as

f_{app} = \frac{Vf}{V - v}                           eq-1

Frequency of moving awayy is given as

f_{aw} = \frac{Vf}{V + v}                          eq-2

Dividing eq-1 by eq-2

\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}

\frac{480}{400} = \frac{343 + v}{343 - v}

v = 31.2 m/s

7 0
3 years ago
A 7 kg sled is initially at rest on a horizontal road. the sled is pulled a distance of 2.9 m by a force of 42 n applied to the
Artemon [7]
I have no idea on this question.
6 0
3 years ago
Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

7 0
3 years ago
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