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Ksivusya [100]
3 years ago
12

In which of the following cases is work being done on an object? Question 2 options: Pushing against a locked door Carrying a bo

x down a corridor Pulling a trailer up a hill Suspending a heavy weight with a strong chain???
Physics
1 answer:
astra-53 [7]3 years ago
3 0

The answer would be:

Pulling a trailer up a hill

Here is why:

In physics, work is computed by the formula:

W = Fd or F(cosΘ)d

Where: F = Force applied on the object

            d =displacement

So in order for work to be done, the object needs to be displaced. In the case of the trailer, the trailer is being pulled and it is travelling in the direction of the force applied.

The other options the object is not displaced. Even if you exert a force, if the displacement remains the same, then it will always be zero, so the force times displacement will be 0, hence no work is done.

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In creating an accurate scale model of our solar system, Lana placed Earth 1 foot from the Sun. The actual distance from Earth t
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3 0
2 years ago
A 1.5-kilogram cart initially moves at 2.0 meters per second. It is brought to rest by a constant net force in 0.30 second. What
AnnZ [28]
Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning.

The cart's acceleration is

                               (0 - 2 m/s) / (0.3 sec)

                           = ( -2 / 0.3 ) (m/s²)  =  -(6 and 2/3) m/s² .

Newton's second law of motion says

                             Force = (mass) x (acceleration) .

For this cart:      Force = (1.5 kg) x ( - 6-2/3 m/s²)

                                       = ( - 1.5 x 20/3 ) (kg-m/s²)

<span>                                       =      </span>- 10 newtons .

<span>The force is negative because it acts opposite to the direction </span>
<span>in which the cart is moving, it causes a negative acceleration, </span>
<span>and it eventually stops the cart.</span>
6 0
3 years ago
A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w
tatyana61 [14]
(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s
And so, the kinetic energy of the object is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s
And so the new kinetic energy is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J

So, the work done on the object is the variation of kinetic energy of the object:
W=\Delta K=120 J-60 J=60 J
7 0
2 years ago
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