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3 years ago

How is amplitude related to loudness

1 answer:

3 0

The loudness of a sound is linked to the size of the vibration which produces it. A big vibration makes a louder sound. Scientists use the word 'amplitude' for the size of waves. For waves on water, it is easy to measure the amplitude.

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Assuming the incline to be frictionless and the zero of gravitational potential energy to be at the elevation of the horizontal

Leya [2.2K]

The energy in the system is given by the law of conservation of energy.

The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h.

Reason:

The given parameters are;

Surface of the inclined plane = Frictionless

Potential energy at the horizontal line = Zero of gravitational potential

By the law of Conservation of Energy, we have;

The spring will be compressed by a distance, x

The energy stored in the compressed spring, K.E. = (1/2)·k·x²

Energy in the block when the block comes to rest at a height, h₁, will be, P.E. = m·g·h₁

Therefore, by conservation of energy, we have;

The initial potential of the block = The stored energy in the compressed string + The gravitational potential energy of the block when it has compressed.

Therefore, the correct option is; The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h

Learn more here;

brainly.com/question/17713698

4 0

2 years ago

Jocelyn is running for the U.S Senate. What is one of the qualifications for office she must meet?

VMariaS [17]

Answer:

Explanation:

4 0

2 years ago

Read 2 more answers

Waves can transfer energy through

lina2011 [118]

Answer:

electromagnetic waves

Explanation:

"wave" is a common term for a number different ways in which energy is transferred

3 0

3 years ago

4. Using the bone density of 2.0 kg/m3, calculate the mass of an adult femur bone that has a volume of 0.00027 m3.

kolbaska11 [484]

Answer:

$\boxed{\sf Mass \ of \ an \ adult \ femur \ bone = 0.00054 \ kg}$

Given:

Bone density = 2.0 kg/m³

Volume of bone (V) = 0.00027 m³

To Find:

Mass of an adult femur bone (m).

Explanation:

$\sf \implies Density = \frac{Mass (m)}{Volume (V)} \\ \\ \sf \implies \frac{Mass}{Volume} = Density \\ \\ \sf \implies Mass = Density \times Volume \\ \\ \sf \implies Mass = 2.0 \ kg/ \cancel{m^{3}} \times 0.00027 \ \cancel{m^{3}} \\ \\ \sf \implies Mass = 0.00054 \ kg$

6 0

3 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.