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denis-greek [22]
3 years ago
13

On an oscilloscope, does a wave with a larger amplitude (greater crests and troughs) indicate louder sound or higher pitch?

Physics
1 answer:
melamori03 [73]3 years ago
8 0
This would be louder sound. Higher pitch would be represented by a more compressed wave that oscillates more times over a shorter period. Higher amplitude represents higher sound because the wave oscillates at the same speed, yet higher.

Hope this helped, and have a nice day!
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an open tank has the shape of a right circular cone (see figure). the tank is 8 feet across the top and 6 feet high. how much wo
Liono4ka [1.6K]

The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.

Given that, the tank is 8 feet across the top and 6 feet high

By the property of similar triangles, 4/6 = r/y

6r = 4y

r = 4/6*y = 2/3*y

Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²

The weight of each disc is m = ρw* A

m = 62.4* 4π/9*y² = 87.08*y²

The distance pumped is 6-y.

The work done in pumping the tank by pumping the water over the top edge is

W = 87.08 ∫(6-y)y² dy

W = 87.08 ∫(6y³ - y²) dy

W =  87.08 [6y⁴/4 - y³/3]

W =  87.08 [3y⁴/2- y³/3]

The limits are from 0 to 6.

W =  87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs

The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.

To know more about work done:

brainly.com/question/16650139

#SPJ4

7 0
2 years ago
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Andrew [12]
The last one, the soil will become weak & unable to support plant growth
4 0
3 years ago
2N2 Hy 03 2N₂²_0₂ +42 H₂O
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4 0
3 years ago
A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (1.40
leva [86]

Answer:

The work done required on the coin during the displacement is 21.75 w.

Explanation:

Given that,

A coin slides over a friction-less plane i.e friction force = 0.

The co-ordinate of the given point is (1.40 m, 7.20 m).

The position vector of the given point is represented by  1.40 \hat i+7.20 \hat j.

The displacement of the coin is

\vec d=1.40 \hat i+7.20 \hat j

The force has magnitude 4.50 N and its makes an angle 128° with positive x axis.

Then x component of the force = 4.50 cos128°

The y component of the force = 4.50 sin128°

Then the position vector of the force is

\vec F=(4.50 cos 128^\circ)\hat i+(4.50 sin 128^\circ)\hat j

   =-2.77 \hat i+3.56 \hat j

We know that,

work done is a scalar product of force and displacement.

W=\vec F.\vec d

    =(-2.77 \hat i+3.56 \hat j).(1.40 \hat i+7.20 \hat j)

    =(-2.77×1.40+ 3.56×7.20) w

    =21.75 w

The work done required on the coin during the displacement is 21.75 w.

6 0
3 years ago
A man weighing 989n on earth only 163 N on the moon. His mass on the moon is __ kg. (Use g=9.8m/s)
shtirl [24]

Answer:

16.63265306 Kg

Explanation:

This is a Newtons to weight conversion.

Fg = weight

Fg = N/9.8

4 0
3 years ago
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