Ask question

Ask question

All categories

4 years ago

15

488 J of work is done to a box which is moved across the floor for a distance of 8.9 m. What net force is required to act on the

box to do this amount of work?

1 answer:

kvv77 [185]4 years ago

8 0

According to the formula
$a = f \times d$
Where a is work, f is force and d is the distance that box was moved over. And from that formula, you can get that f = a/d and that is 54.83N of force

You might be interested in

The power in an electrical current is given by the equation

Tcecarenko [31]

Answer:

P = VI

Explanation:

the power is equal to the current × voltage

6 0

2 years ago

Read 2 more answers

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

If a fish experiences an a of 9.8m/s2 over a period of 1.84s, what was the 4v?

777dan777 [17]

Answer:

It’s 18.0 m/s

Explanation:

Use acceleration formula then plug in 9.8 and 1.84s

6 0

3 years ago

Convert 35 centimeters to inches

SIZIF [17.4K]

Answer:

13.7795276 Inches

Explanation:

5 0

3 years ago

The scientist who suggested that energy can be created under certain conditions was

tester [92]

Einstein hope this helped

7 0

4 years ago

Read 2 more answers