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saul85 [17]
4 years ago
15

true or false Newtons's First Law states "Acceleration depends on the mass of the object and the unbalanced force applied"

Physics
2 answers:
babunello [35]4 years ago
8 0
False, his first law states: An object that's in motion will remain in motion at a constant velocity unless it's acted on by an unbalanced force.

Nady [450]4 years ago
7 0
The answer to your question would be False. Newton's First Law actually states "a body at rest (or in motion) will remain at rest (or in motion) until acted upon by an unbalance force."
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A pump lifts 400 kg of water per hour a height of 4.5 m .
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3 years ago
A tank, shaped like a cone has height 6 meters and base radius 5 meters long. It is placed so that the circular part is upward.
AlekseyPX

The work required to pump out the liquid is 2.35 MJ.

What is work?

Work can be defined as the scalar product of the force and the displacement.

The formula to calculate the work done W is,

W=F.d

where F is the force and d is the displacement.

Note: It is assumed that the weight of one cubic unit of the liquid is 10000 \text{ N/m}^3.

Break down the vertical tank into parallel disc elements whose thickness is dh and are located at a height h from the ground. The radius of the element at height h is r.

It is given that the height of the tank is 6 m and the base radius is 5 meters. Using these, the ratio of the radius r at height h to the height h by using the triangle similarity criterion is,

\frac{h}{r}=\frac{6}{5} \\ r=\frac{5h}{6}

Using the above value of r, the volume of the element is given by,

V= \pi r^2dh\\V=\pi (\frac{5h}{6})^2dh\\V= \frac{25\pi h^2}{36} dh

The weight of the volume element is the force required.

The weight of one cubic unit of the liquid is 10000 \text{ N/m}^3.

So the weight F of the volume element at height h is given by,

F=10000 (\frac{25 \pi h^2}{ 36})dh\\F= 6944.44\pi h^2dh

At the height h, the difference in the height of the top of the tank and the volume element is (6-h). So the work required to pump out the volume of liquid at height h is given by the integrating the equation,

W=6944.44\pi\int\limits^6_0 {(6h^2-h^3)} \, dx \\\\W=69.44\pi \left[ 2h^3-\frac{h^4}{4}\right]^6_0\\W=2,.35\times10^6\text{ J}

The value of the W obtained is 2.35*10^6 J or 2.35 MJ.

Learn more about work here:

brainly.com/question/16627899

#SPJ4

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2 years ago
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