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Kaylis [27]
3 years ago
11

An electric bell is kept under a bell jar in which a vacuum has been created. Slowly air is introduced in to the jar. What will

happen to the sound of the electric bell?Select one of the options below as your answer:. A. It will increase until it reaches its maximum.. B. It will decrease until it stops.. C. It will not change.. D. It will increase and decrease periodically. .E. It will decrease exponentially
Physics
2 answers:
Aleks [24]3 years ago
8 0
A. Increase until it reaches it's maximum. Sound travels through air and as the amount of air in the jar increases, the loudness or amplitude of the bell increases until it reaches it's maximum at normal air pressure.
ser-zykov [4K]3 years ago
4 0

The sound of the electric bell will increase until it reaches its maximum. The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query about and it will be able to help you, and if you’d like, feel free to ask another question.

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When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that typical
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Answer:

Explanation:

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Frequency is defined as the number of oscillations or vibrations completed in one second.

The reciprocal of time period is called frequency.

f = 1 / T

f = 1 / (50 x 10^-6)

f = 20,000 Hz

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7 0
3 years ago
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A 1.0 kg ball at the end of a 2.0 m string swings in a vertical plane. At its lowest point the ball is moving with a speed of 10
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Answer:

a) 4.65m/s

b) 59.8 N , 1.01125 N

Explanation:

a)

m = mass of the ball = 1 kg

r = length of the string = 2.0 m

h = height gained by the ball as it moves from lowest to topmost position = 2r = 2 x 2 = 4 m

v = speed at the lowest position = 10 m/s

v' = speed at the topmost position = ?

Using conservation of energy

Kinetic energy at topmost position + Potential energy at topmost position = Kinetic energy at lowest position

(0.5) m v'² + m g h = (0.5) m v²

(0.5) v'² + g h = (0.5) v²

(0.5) v'² + (9.8 x 4) = (0.5) (10)²

v' = 4.65m/s

b)

T' = Tension force in the string when the ball is at topmost position

T = Tension force in the string when the ball is at lowest position

At the topmost position:

force equation is given as

mg + T' = \frac{m v'^{2}}{r}

(1)(9.8) + T' = \frac{(1) (4.65)^{2}}{2}

T' = 1.01125 N

At the lowest position:

force equation is given as

T - mg = \frac{m v^{2}}{r}

T - (1) (9.8) = \frac{(1) (10)^{2}}{2}

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It is said to be concave
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Anwser

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Answer:

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