All categories

2 years ago

The container was lifted a height of 14 m

Physics

1 answer:

Alika [10]2 years ago

8 0

Work done = Force X Distance

3 430 000J = Force X 14m

Force = 3 430 000J / 14m
= 245 000 N

Hope this helps!

You might be interested in

A gravitational _____ exists between you and every object in the universe.

andrew-mc [135]

A pair of equal gravitational forces ... one in each direction ...
exists between every speck of mass in the universe and every
other speck of mass.

4 0

3 years ago

Read 2 more answers

A single point charge is placed at the center of an imaginary cube that has 10 cm long edges. The electric flux out of one of th

12345 [234]

Answer:

Explanation:

<u>Given:</u>

Length of each side of the cube, $L=10\ cm$

The Elecric flux through one of the side of the cube is, $\phi =-1 kNm^2/C.$

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by $\epsilon_0$

Since Flux is a scalar quantity. It can added to get total flux through the surface.

$\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C$

So the the charge at the centre is calculated.

5 0

3 years ago

José and Laurel measured the length of a stick's shadow during the day. Without knowing the length of the stick, which of their

skelet666 [1.2K]

Answer:

Option B.

Explanation:

Assuming the stick is in vertical position, its shadow depends on two factors: its length and the angle between the sun rays and the stick. When the angle is bigger, the lenght of the shadow increases, and vice versa. So, when the sun rays are parallel to the stick, the shadow may be small. Since they are nearly perpendicular to the Earth's surface at 12 o'clock, the shadow of the stick at that time should be minimal. It means that the measured shadow of 75 cm at 12:30 p.m. is almost impossible (Option B).

5 0

3 years ago

You are in Paris, 60 m up in the Eiffel Tower. If you throw a euro downward at a velocity of 2.0 m/s, how long would it take the

kondor19780726 [428]

Answer:

t = 3.29 seconds

Explanation:

It is given that,

Height of the Eiffel tower is 60 m

Initial speed of a euro, u = 2 m/s

It will move under the action of gravity in the downward direction. Firstly, we can find the final velocity as follows :

$v^2-u^2=2ad\\\\v=\sqrt{u^2+2ad} \\\\v=\sqrt{(2)^2+2\times 9.81\times 60} \\\\v=34.36\ m/s$

Let t is the time taken by the euro to hit the ground. It can be calculated as :

$v=u+at\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{34.36-2}{9.81}\\\\t=3.29\ s$

Hence, it will take 3.29 seconds to hit the ground.

4 0

3 years ago

What major region does this profile<br> most likely represent?

alexdok [17]

Answer:

A mid ocean ridge possibly a plate margin spreading area

Explanation:

6 0

3 years ago