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Dennis_Churaev [7]
3 years ago
10

You need to fill a basketball with air to play with it. You know that your pump expels air at a velocity of 6 cm/s. You know tha

t the volume flow rate of air being pumped into the basketball is 103 cm^3/s. What is the diameter of your needle?
Physics
1 answer:
Scrat [10]3 years ago
8 0

Answer:

The diameter of the needle is <u>4.675 cm</u>.

Explanation:

Given:

Volume flow rate is, \dot V=103\ cm^3

Velocity of air expelled by pump is, v=6\ cm/s

Let the area of the needle be 'A' cm² and the diameter be 'd' cm.

We know that, volume flow rate of the air expelled by pump is given as the product of the needle's area and velocity of air flowing through that area.

Therefore, volume flow rate is given as:

\dot V=A\times v\\\\103=A\times 6\\\\6A=103\\\\A=\frac{103}{6}\ cm^2

Now, considering the needle to be circular, area of the needle can be written as:

A=\frac{\pi}{4}\times d^2\\\\\frac{103}{6}=\frac{\pi}{4}\times d^2\\\\d^2=\frac{103\times 4}{6\pi}\\\\d=\sqrt{\frac{103\times 4}{6\pi}}\\\\d=4.675\ cm

Therefore, the diameter of the needle is 4.675 cm.

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A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and t
vladimir2022 [97]

This question is incomplete, the complete question is;

Flag

A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).

Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.

Answer:

the work done is -88 J

Explanation:

Given the data in the question;

we know that;

Work done = F × S

where constant force F = ( 6i + 8j - 6k )

S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )

S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )

S = ( -12I + 7j + 12k )

so

Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )

Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )

Work force = -72 + 56 - 72

Work force = -88 J

Therefore, the work done is -88 J

8 0
3 years ago
g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller
levacccp [35]

Answer:

a) v₂ = 30 m/s

b) m₁ = 12600 kg

c) m₂ = 12600 kg

Explanation:

a)

Using the continuity equation:

A_1v_1 = A_2v_2

where,

A₁ = Area of inlet = π(0.15 m)² = 0.07 m²

A₂ = Area of outlet = π(0.05 m)² = 0.007 m²

v₁ = speed at inlet = 3 m/s

v₂ = speed at outlet = ?

Therefore,

(0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}

<u>v₂ = 30 m/s</u>

<u></u>

b)

m_1 = \rho A_1v_1t

where,

m₁ = mass of water flowing in = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\

<u>m₁ = 12600 kg</u>

<u></u>

c)

m_1 = \rho A_1v_1t

where,

m₂ = mass of water flowing out = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\

<u>m₂ = 12600 kg</u>

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