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3 years ago

A force of 350N is applied to a body. If the work done is 40kJ, what is the distance through which the body moved?

Physics

1 answer:

Studentka2010 [4]3 years ago

7 0

The distance covered by the body is 114.3 m

Explanation:

The work done by a force exerted on an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the object in this problem, we have

F = 350 N is the force applied

$W=40 kJ = 40,000 J$ is the work done

$\theta=0^{\circ}$ if we assume that the force is applied parallel to the motion of the object

Solving for d, we find the distance covered by the object:

$d=\frac{W}{F cos \theta}=\frac{40,000}{(350)(cos 0)}=114.3 m$

Learn more about work:

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Question 11 (1 point)

Ainat [17]

Answer:

gravitational potential energy.

Explanation:

Gravitational potential energy (GPE) can be defined as an energy possessed by an object or body due to its position above the earth surface.

Mathematically, gravitational potential energy is given by the formula;

$G.P.E = mgh$

Where,

G.P.E represents gravitational potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

This ultimately implies that, anytime there is height, the object must have gravitational potential energy.

Hence, an object possesses gravitational potential energy due to its height (position) and the earth's gravitational force.

8 0

3 years ago

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

3 years ago

Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many

Zielflug [23.3K]

Answer:

The $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

Explanation:

Given :

Current $I = 1 \times 10^{5}$ A

Charge of electron $e = 1.6 \times 10^{-19}$ C

Time $t = 1$ sec

From the formula of current,

Current is the number of charges flowing per unit time.

$I = \frac{ne}{t}$

Where $n =$ number of charges means in our case number of electrons

$n = \frac{It}{e}$

$n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }$

$n = 6.25 \times 10^{23}$

Therefore, $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

5 0

3 years ago

Determine the amount of potential energy of a 5N book that is 1.5m high on a shelf.

Alex73 [517]

Answer:

Potential energy of book = 7.5 J

Explanation:

Given:

Weight of book = 5 N

Height of shelf = 1.5 meter

Find:

Potential energy of book

Computation:

Weight = Mass x Acceleration of gravity

Mass x Acceleration of gravity = 5 N

Potential energy = Mass x Acceleration of gravity x Height

Potential energy of book = Mass x Acceleration of gravity x Height

We know that;

Mass x Acceleration of gravity = 5 N

So,

Potential energy of book = 5 x 1.5

Potential energy of book = 7.5 J

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3 years ago

The picture below shows a wheelbarrow. Using a wheelbarrow can make it easier to lift a heavy object by

timofeeve [1]

Answer:

B IS CORRECT

Explanation:

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