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svetoff [14.1K]
3 years ago
8

What term describes how far an object moves in a certain amount of time

Physics
1 answer:
stepan [7]3 years ago
8 0

Answer:

1. Speed

Explanation:

"how far an object moves in a certain amount of time"

"how far an object moves"- <em>distance</em>

"In a " - <em>over</em>

"certain amount of time" - <em>time</em>

<em>Distance over time is speed. </em>

The speed of an object is the distance the object travels in one unit of time.

<em>Hope I Helped, Feel free to ask any questions to clarify :)</em>

<em>Have great day!</em>

<em>            -Aadi x</em>

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When the government sets a price for wheat that is above the equilibrium price, it is imposing a _____.
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<h3><u>Answer</u>;</h3>

Price floor

When the government sets a price for wheat that is above the equilibrium price, it is imposing a<u> price floor</u>.

<h3><u>Explanation</u>;</h3>
  • <u>Price floors prevent a price from falling below a certain level. When a price floor is set above the equilibrium price, quantity supplied will exceed quantity demanded, and excess supply. </u>
  • <u>Price control </u>occurs when the government laws regulate prices instead of letting market forces determine prices.
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Is an aquarium a system ?
sweet [91]

No

Explanation:Brooo it does not contribute to the eco system

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The atomic mass of an atom do not contain electrons because
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Answer:

Electrons are so small that it does not affect the mass of atom .

Explanation:

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7 0
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A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is
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Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12  m

Let assume the tank have a thickness  = dh

The radius of the tank by using the concept of similar triangle is :

\dfrac{1}{r} = \dfrac{12}{h}

r = \dfrac{h}{12}

The area of the tank = \mathbf{\pi r^2}

The area of the tank = \mathbf{\pi( \dfrac{h}{12})^2}

The area of the tank = \mathbf{ \dfrac{\pi}{144}h^2}

The volume of the tank is  = area × thickness

= \mathbf{ \dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \rho_ g * volume

where;

\rho_g = density of water ; which is given as 10000 N/m³

So;

Weight of the element = \mathbf{ 10000 *\dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \mathbf{69.44 \ \pi  \ h^2 \  dh}

However; the work required to pump this water = weight × height  rise

where the height rise = 12 - h

the work required to pump this water  = \mathbf{69.44 \ \pi  \ h^2 \  dh}(12 - h)

the work required to pump this water  = \mathbf{69.44 \pi (12h^2-h^3)dh}

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = \mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}

= \mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}

=  \mathbf{ 69.44 \pi[ \frac{12h^3}{3}-  \frac{h^4}{4}]^{12}}_0} }

= \mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}

= \mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}

= \mathbf{69.44 \pi*12^4 *\frac{1}{12}}

= 376966.991 Joules

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