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3 years ago

In which situations is gravitational potential energy present? Check all that apply.

2 answers:

8 0

A person walks up a flight of stairs √
Wind lifts a balloon into the air √
A weightlifter holds a barbell straight overhead √

Gravity is used in most real world examples because if not of gravity everyone would be floating all around ^-^

Hope this helps ;p

Arte-miy333 [17]3 years ago

8 0

Answer:

Option (1), (4), (6)

Explanation:

The energy possessed by the position or configuration of a body is called it's potential energy.

Gravitational potential energy is defined as the energy possessed by the height of object from the ground level.

Here, a person walks up on the flight of stairs has cover some height from the ground level, so it has gravitational energy.

Wind lifts a balloon in air also have gravitational potential energy.

Weightlifter lifts a barbell overhead also cover some height to lift the barbell, so it also have some gravitational potential energy.

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An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi

shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = $a^{2}$

Let the height of the bin be 'h'

Therefore the total area, $A_{t} = 4ah$

The cost is:

C = 2sh

Volume of the box, V = $a^{2}h = 4^{2}\times 2 = 128$ (1)

Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

$C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}$

$C(a) = 2a^{2} + \frac{256}{a}$

Differentiating the above eqn w.r.t 'a':

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

a = 4

Also

$h = \frac{128}{4^{2}} = 8$

The path in order to minimize the cost must be a rectangle.

8 0

4 years ago

Does the image above correctly illustrate how the coriolis effect impacts global winds?

Sedbober [7]

Yes the winds are moving in a straight line

6 0

3 years ago

A student that jumps a vertical height of 50 cm during the hang time activity.

muminat

The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s

Why?

To solve the problem, we must consider the vertical height reached by the student as max height.

We can use the following equations to solve the problem:

Initial speed calculations:

$v_{f}^{2}=v_{o}^{2}+2*a*d$

At max height, the speed tends to zero.

So, calculating, we have:

 $v_{f}^{2}=v_{o}^{2}+2*a*d\\\\0=v_{o}^{2}+2*(-9.81\frac{m}{s^{2}})*0.5m\\\\v_{o}^{2}=9.81\frac{m^{2} }{s^{2}}\\\\v_{o}=\sqrt{9.81\frac{m^{2} }{s^{2}}}=3.13\frac{m}{s}$ 

Hang time calculations:

We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:

$y-yo=vo.t+\frac{1}{2}*9.81\frac{m}{s^{2}}*t^{2} \\\\0.5m-0=0*t+4.95*t^{2}\\\\t^{2}=\frac{0.5}{4.95}\\\\t=\sqrt{0.101}=0.318s$