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3 years ago

11

If a ball has the weight of 3.76, how much work would it take to lift it 2 meters above the ground? (sorry if this question does

n't have all the components, im bad with physics)

1 answer:

Aloiza [94]3 years ago

7 0

We know, work done = Weight * Displacement
w = 3.76 * 2
w = 7.52 J

In short, Your Answer would be: 7.52 Joules

Hope this helps!

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<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :

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As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

$X=V_{ox}.t$ (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$ , we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$ (6)

$t=2.42s$ (7)

<h2>In the y-axis: </h2>

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$ (8)

Where:

$y$ is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$ (9)

$-y=-10.52m$ (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building

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A ball is dropped from a tower that is 206 m high, How long does it take to reach the ground?

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From the question given above, the following data were obtained:

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The time taken for the ball to get to the ground can be obtained as follow:

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Learn more: brainly.com/question/24903556

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