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4 years ago

We measure the loudness of sound in decibels. a. True b. False

Physics

1 answer:

Tatiana [17]4 years ago

7 0

Answer: The correct answer is True.

Explanation:

Loudness of sound is referred to how soft or loud a sound is for the listener.

This term is measured in a unit known as decibels referred to as dB.

This unit is used to measure the relative intensity of sounds on a scale from zero to 100 dB.

More the value of decibels, it will be uncomfortable for a person to hear that sound.

So Yes, the loudness of sound is measured in decibels.

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If a force does a positive amount of work on an object, does the object’s speed increase, decrease, or remain the same? Justify

iris [78.8K]

Answer:

see that there is no dependence on speed, so the work remains constant

Explanation:

Work is defined by the expression

W = F. d

where the boldface indicates vectors, this equation can be written in scalar form

W = f d cos θ

where θ is the angle between force and displacement.

We see that there is no dependence on speed, so the work remains constant

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3 years ago

Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius o

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Answer:

the distance of image from the vertex is 45 cm and the image formed is in the glass.

Explanation:

distance of object, u = - infinity

radius of curvature, R = - 15 cm

refractive index, n = 1.5

Let the distance of image is v.

Use the formula

$-\frac{n1}{u}+\frac{n2}{v}=\frac{n2- n1}{R}\\\\-\frac{1}{\infty }+\frac{1.5}{v}=\frac{1.5-1}{-15}\\\\v=45 cm$

The image is in the glass.

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3 years ago

A person hits a 45-g golf ball. The ball comes down on a tree root and bounces

Verizon [17]

Answer:

Maximum height, h = 10 m

Explanation:

It is given that,

Mass of golf ball, m = 45 g = 0.045 kg

The ball comes down on a tree root and bounces straight up with an initial speed of 14.0 m/s.

We need to find the height the ball will rise after the bounce. It is based on the conservation of energy such that,

$\dfrac{1}{2}mv^2=mgh$

h is maximum height raised by the ball

$h=\dfrac{v^2}{2g}\\\\h=\dfrac{(14)^2}{2\times 9.8}\\\\h=10\ m$

So, the ball will raised to a height of 10 meters.

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The fundamental mode of vibration of a wave is defined as that which has the LOWEST frequency

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