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Sedaia [141]
2 years ago
5

A shop-vac is capable of pulling in air at a rate of 210 ft^3/min. what is the rate of the vacuum's air flow in l/s?

Physics
1 answer:
Nadya [2.5K]2 years ago
3 0

The correct answer is 99.06 .

Since we know that,

1 ft = 12 inches, (1ft)3 = (12)3inches3

& 1 inch = 2.54 cm ,  (1 inch)3 = (2.54)3cm3

& 1 cm3 = 1ml,

&(1 ml)/1000 = 1L, so (1cm3)/1000 = 1L.

& 1 min. = 60 sec.

Now, 1ft3 = (12)3x(2.54)3/(1000) L

and 1 min = 60 sec

So, 1ft3/min = (12)^3x(2.54)^3/(1000)x(60) L/s

Hence 210 ft3/min = 210x(12^)3x(2.54)^3/(1000)x(60).

= 99.06.

Learn more about rate of the vacuum's air flow here :-

brainly.com/question/24641042

#SPJ4

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The latent heat of fusion refers to the solid to liquid or liquid to solid states.

Answer: Option C

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This external energy should be greater than the latent heat of solid in order to successfully break the bonds to form liquid. So the change in the enthalpy of the reaction while conversion from solids to liquids are termed as latent heats of fusion.

Even the inter-conversion from liquid to solid state will undergo change in enthalpy where the heat will be released and that is termed as latent heats of solidification. It is found that latent heat of solidification is equal in magnitude but opposite in direction with the latent heats of fusion.

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Answer:

(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N

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(3) The period of a satellite is given as;

T = 2\pi \sqrt{\frac{r^3}{GM} }

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T is the period of the satellite

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r is the radius of the orbit

Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.

 

(4)

Given;

mass of the ball, m₁ = 1.99 x 10⁴⁰ kg

mass of Neptune, m₂ = 1.03 x 10²⁶ kg

mass of Sun, m₃ = 1.99 x 10³⁰ kg

distance between the Sun and Neptune, r = 4.5 x 10¹² m

The gravitational force between the Sun and Neptune is calculated as;

F_g = \frac{Gm_2m_3}{r^2} \\\\F_g = \frac{6.67\times 10^{-11} \times 1.03 \times 10^{26}\times 1.99\times 10^{30}}{(4.5\times 10^{12})^2} \\\\F_g = 6.751 \times 10^{20} \ N

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2.Atomic bonds

3.Compound

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Answer: 2.55meter

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