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bixtya [17]
3 years ago
5

You and a friend are playing tug-of-war with a massless rope. You are pulling with 50 Newtons of force while your friend is pull

ing with 40 Newtons of force.
What is the tension in the rope?
What is your net force?
What is your friend’s net force?
Physics
1 answer:
const2013 [10]3 years ago
4 0

Answer:

Explanation:

A tug of war is game where two teams will be pulling a rope horizontally,

So,

Given that,

You applied a force of F1 = 50N

You friend you are pulling against applied a force of F2 = 40N

A. Tension on the rope.

Let the tension on the rope be T

The tension cause by you on the rope is the force you applied

T = 50N.

The tension cause by your friend is also

T = 40N

Adding the two tension together

T+T = 50+40

2T = 90

T = 45

OR you will see the rope as a pulley system, i.e. we will assume that the two friend force cause a tension on the rope, their are two tension on the rope, and two forces are applied

Then, tension is equal to force applied

T+T = 40+50

2T = 90

T = 45N

b. Net force by you

In this case we will take your position as the positive direction, I.e. the net force cause by you

Using newton second law

Fnet. = ΣFx

Fnet = F1—F2

Fnet = 50—40

Fnet = 10 N

c. Fnet by your friend

In this case we will take your friends position as the positive direction, I.e. the net force cause by your friend.

Fnet. = ΣFx

Fnet = F2—F1

Fnet = 40—50

Fnet = —10 N

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True or False. The pressure inside a piston is the same in all directions.
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4 0
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Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di
jeyben [28]

<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at 20^0\ C is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

d=\dfrac{\lambda}{2}........(1)

Velocity of sound wave is given by :

v=f\times \lambda

d=\dfrac{v}{2f}

d=\dfrac{343\ m/s}{2\times 686\ Hz}

d=0.25\ m

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

d=n\lambda  ( n = integers )

Let n = 1

So, d=\dfrac{v}{f}

d=\dfrac{343\ m/s}{686\ Hz}

d=0.5\ m

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

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