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Arturiano [62]
3 years ago
10

A Russian athlete lifts 300kg to a height of 2.5 m in 2 seconds. What is the work done? And what power is exerted?

Physics
1 answer:
snow_tiger [21]3 years ago
5 0
We want to find the work done and power exerted, let’s start with work first.

We know that the equation for work is: W = F * D. We need to find the force which we can find by using: F = M * A.

Mass: 300kg
Acceleration (using equation from photo): 1.25 m/s^2

(The equation says x but can be used with y values)

If you are confused about how I found the acceleration; I plugged in 2.5 for the final y value, 0 for the initial y value, 0 for the initial velocity and 4 for t squared.

To solve, for acceleration it’s a matter of simple algebra. You can subtract the initial y position and the initial velocity from the final y position because they are 0. This leaves you with 2.5 m = 1/2a * t^2, from here I multiplied 2.5 by 2 to get rid of the 1/2. Now I have 5 = a * t^2. T^2 is just 2 squared, so four. Simply divide 5 by 4, and boom, you get 1.25 m/s^2.

Force = 300 kg * 1.25 m/s^2 = 375 Newtons

So, work = 500 N * 2.5 m = 1000 Joules

Power: W/t

So, Power = 1000 J / 2 seconds = 500 Watts

Hope this helps!

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The spinning of the earth around its own axis causes day and night.

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Answer:  Aerial plants is plants that lives in air or wind the wind serves as the water of the plants. Most aerial plants are found in tropical and equatorial regions of the world. In evergreen rain forests, the foliage is so thick that some plants have evolved aerial roots to allow them to absorb more sunlight. The development of aerial roots is

thus an evolutionary process.Aerial roots are often thick and spread around the parent tree. The Banyan tree can have several aerial roots as it gets older.

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7 0
3 years ago
A brick falls to the ground. if the time for the collision of the brick and the ground is increased by a factor of 4, the force
melomori [17]

Answer:

By a factor of 1/4.

Explanation:

The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,

\\\begin{aligned}\\\small F &=\small \frac{\Delta (mV)}{\Delta T}\end{aligned}

in which \small \Delta (mV) , \small \Delta t represent the change in momentum and the time taken for that change.

If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by \small F_1, the following manipulation confirms the answer to this question.

\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}

Here \small F is the force that was applied to the object previously.

#SPJ4

4 0
2 years ago
A toy rocket is launched straight up by using a spring. The rocket is initially pressed down on the spring so that the spring is
tigry1 [53]

Answer:

6.86 meters

Explanation:

Let the compression of the string be represented by x, and the height of projection of the toy rocket be represented by h.

So that;

x = 9 cm = 0.09 m

In its rest position (i.e before the launch), the spring has a stored potential energy which is given as;

PE = \frac{1}{2} Kx^{2}

    = \frac{1}{2} x 830 x (0.09)^{2}

    = 415 x 0.0081

    = 3.3615

The potential energy in the string = 3.36 Joules

Also,

PE = mgh

where: m is the mass, g is the gravitational force and h the height.

m = 50 g = 0.05 kg, g = 9.8 ms^{-2}

Thus,

PE = 0.05 x 9.8 x h

3.3615 = 0.05 x 9.8 x h

3.3615 = 0.49h

⇒ h = \frac{3.3615}{0.49}

      = 6.8602

The height of the toy rocket would be 6.86 meters.

4 0
3 years ago
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