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3 years ago

Okay, I have three more questions.

2 answers:

5 0

When a liquid or a gas is heated, it expands and becomes less dense, so it rises, while the cooler, denser liquid or gas sinks. (Not sure if this helps but this is what I hit)

Solnce55 [7]3 years ago

4 0

1. Less
2. More
3. Convection current
I am unsure if 3 is correct

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How many light years away is the sun from the middle of the Millky way

julsineya [31]

Answer:

The Milky Way is about 1,000,000,000,000,000,000 km (about 100,000 light years or about 30 kpc) across. The Sun does not lie near the center of our Galaxy. It lies about 8 kpc from the center on what is known as the Orion Arm of the Milky Way

4 0

3 years ago

An air-filled pipe is found to have successive harmonics at 700 Hz , 900 Hz , and 1100 Hz . It is unknown whether harmonics belo

Artyom0805 [142]

Answer:

the length of the pipe is 0.85 m or 85 cm

Explanation:

Given the data in the question;

The successive harmonics are; 700 Hz , 900 Hz , and 1100 H

Now, for a closed pipe,

length of pipe (L) = λ/4

Harmonics; 1x, 3x, 5x, 7x, 9x, 11x

1100Hz - 900Hz = 200Hz

⇒ 2x = 200Hz

x = 100Hz ( fundamental frequency )

λ = V/f = 340 /100 = 3.4 m

Now

Length L = λ / 4

L = 3.4 / 4

L = 0.85 m or 85 cm

Therefore, the length of the pipe is 0.85 m or 85 cm

7 0

3 years ago

A little girl riding a train rolls a ball toward the back of the train at 1.25 m/s NE. The train is traveling at a velocity of 1

Lelechka [254]

Answer:

Answer: 2.70m/s NE

Explanation:

Just did it.

4 0

4 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

4 years ago

An object of mass 20 kg is raised vertically through a distance of 8 m above ground level. If g = 10 m/s2, what is the gravitati

IgorC [24]

2400joules

Explanation:

P.E

m= 20kg h=8m g=10m

P.E= 20×8×10

=2400joules

6 0

2 years ago