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3 years ago

How does increasing the starting height affect the final speed?

Physics

1 answer:

Lady bird [3.3K]3 years ago

7 0

Answer:

To investigate into how the height of a ramp affects the speed of a toy car rolled down a hill or slope because the higher the ramp the more Gravitational potential Energy (GPE), the more kinetic energy converted and the car will go faster. So the ramp increases, the velocity will also increase.

Explanation:

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A container of gas is at a pressure of 3.7 x 10^5 Pa. How much work is done by the gas if its volume expands by 1.6 m^3 ?

Dmitriy789 [7]

Answer:

592000 J

Explanation:

We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:

1 Pa = 1 Kg/ms²

Therefore,

3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²

Next, we shall determine the workdone.

Workdone is given by the following equation:

Workdone (Wd) = pressure (P) × change in volume (ΔV)

Wd = PΔV

With the above formula, the work done can be obtained as follow:

Pressure (P) = 3.7×10⁵ Kg/ms²

Change in volume (ΔV) = 1.6 m³

Workdone (Wd) =?

Wd = PΔV

Wd = 3.7×10⁵ × 1.6

Wd = 592000 Kgm²/s²

Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:

1 Kgm²/s² = 1 J

Therefore,

592000 Kgm²/s² = 592000 J

Therefore, the Workdone is 592000 J.

6 0

3 years ago

The turbines can be seen inside this hydroelectric dam. Why are they located at that particular height?

Yakvenalex [24]

Answer:

Explanation:

the answer is number three

5 0

4 years ago

Read 2 more answers

a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on

alekssr [168]

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

$F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}$

Now, if the body is on the surface of the Earth, its weight (w) would be:

$F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}$

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

$\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\$

4 0

3 years ago

A light body and a heavy body have the

Elena-2011 [213]

Answer:

The heavy body

Explanation:

The heavy body because it's heavier and the heavier something is the more kinetic energy it has.

Pretty sure this is correct

Hope it helps

7 0

3 years ago

Read 2 more answers

A metal rod has a length of 123. cm at 200°C. At what temperature will the length be 92.6 cm if the coefficient of linear expans

mestny [16]

Answer:

$\theta_{2} = 15400^0 C$

Explanation:

The formula for linear expansivity, $\alpha = \frac{l_{2} - l_{1} }{l_{1} ( \theta_{2} - \theta_{1} )}$

original length, l₁ = 123 cm = 1.23 m

final length, l₁ = 92.6 cm =0.926 m

original temperature, θ₁ = 200°C

Linear expansivity, α = 2 * 10⁻⁵ °C⁻¹

Putting these values into the formula:

$2 * 10^{-5} = \frac{1.23 - 0.926 }{l_{1} ( \theta_{2} -200 )}\\ \theta_{2} -200 = \frac{0.304}{2 * 10^{-5} } \\\theta_{2} = 15200 + 200\\\theta_{2} = 15400^0 C$

7 0

3 years ago