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IRISSAK [1]
3 years ago
15

What is a measure of the pulling force of gravity?

Physics
1 answer:
asambeis [7]3 years ago
7 0

Question asked:What is a measure of the pulling force of gravity?

My answer: newtons

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One student bangs two bricks together
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Answer:

This could be done if a stop watch is used to calculate the time taken to hear the echo and a rule should be used to calculate the distance between the bricks and the wall. Then divide distance by time

Explanation:

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7 0
4 years ago
Câu 1. Trường hợp nào dưới đây không phải là vật sáng?
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Answer:

A

Explanation:

A. The pencil is on the table in broad daylight

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3 years ago
Which of the following best describes what alveolar are
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Alveoli are tiny balloon shaped structures and are the smallest passageway in the respiratory system. The alveoli are only one cell thick, allowing the relatively easy passage of oxygen and carbon dioxide (CO2) between the alveoli and blood vessels called capillaries.
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In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fitted. The head of water on the meter wh
horrorfan [7]

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Q=C_d\frac{A_1A_2}{\sqrt{A_1^2-A_2^2}}\sqrt{2gh}

Where,

C_d is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ = \frac{\pi}{4}0.1^2 = 7.85\times 10^{-3}m^2

A₂ = Area at the throat

A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

5 0
3 years ago
Which, if any, of the following statements about electric field lines is/are true? A.The electric field is always perpendicular
jonny [76]
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If a proton were placed on an electric field line, it would move in a direction anti-parallel to the field.
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8 0
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