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serg [7]
3 years ago
11

15) In which of the following cases would sound reach each ear out of phase? A. You are standing directly in front of the sound

source. B. You are standing directly below the sound source. C. You are standing directly to the right of the sound source. D. You are standing directly behind the sound source.
16) In which of the following cases would sound reach each ear out of phase? A. You are standing directly in front of the sound source.
B. You are standing directly below the sound source.
C. You are standing directly to the right of the sound source.
D. You are standing directly behind the sound source.
Physics
2 answers:
Sonja [21]3 years ago
7 0

In one of the greatest coincidences to arise on Brainly in quite some time, Choice-C is the correct choice for BOTH #15 and #16 .

Choice-C is the only situation in which the source is a different distance from each ear.

Oksana_A [137]3 years ago
5 0

Answer:

(15)The sound will reach ear out of phase when you are standing directly to the right of sound source.

Because when we are behind or in front distance between both the ear to the sound source will be same.

Therefore, Option C is correct.

(16) The sound  source will reach ear out of phase when you are standing directly to the right of sound source.

Therefore, option C is correct.

Both questions have same information.

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D. The sum total of all matter and energy

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A cart of mass M = 2.40 kg can roll without friction on a level track. A light string draped over a light, frictionless pulley c
sergiy2304 [10]

Answer:

Part a)

a_{hanger} = g - \frac{T}{m}

Part b)

a_{cart} = \frac{T}{M}

Part c)

a = \frac{mg}{M + m}

Part d)

a = 1.35 m/s^2

Explanation:

Part a)

For hanger we know that it will have tension force upwards while it has downwards its weight so we will have

mg - T = ma

so we have

a_{hanger} = g - \frac{T}{m}

Part b)

now for car that is rolling on the floor the net force is given as

F = Ma

T = Ma

a_{cart} = \frac{T}{M}

Part c)

now we know that the cart and the hanger both are connected to each other

so they must have same acceleration

so we will have

T = Ma

mg - Ma = ma

a = \frac{mg}{M + m}

Part d)

now we know that

M = 2.40 kg

m = 0.50 kg

so we will have

a = \frac{0.50(9.81)}{2.40 + 0.50}

a = 1.35 m/s^2

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