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Paraphin [41]
3 years ago
7

A box has a momentum of 38.0 kg*m/s to the right. A 88.3 N force pushes it to the right for 0.338 s. What is the final momentum

of the box? (unit = kg*m/s)
Physics
2 answers:
bixtya [17]3 years ago
4 0

Answer:

67.8

Explanation:

im an acellus student, and this is the answer that i got when doing the problem.

lana [24]3 years ago
3 0

Answer:

63.8454kgm/s

Explanation:

According to Newton's second law;

Force = mass ×acceleration

Fore = m(v-u)/t

Force = mv-mu/t

MV is the final momentum

mu is the initial momentum = 38kgm/s

Force = 88.3N

Time = 0.338s

Substitute and get mv

88.3 = mv-38/0.338

88.3×0.338 = mv-38

29.8454 = mv-38

MV = 29.8454+34

mv = 63.8454kgm/s

Hence the final momentum is 63.8454kgm/s

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How far away is a cliff if an echo is heard 0.486 s after the original sound? Assume that sound traveled at 343 m/s on that day.
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4 0
2 years ago
A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud
klio [65]

If the applied force is in the same direction as the object's displacement, the work done on the object is:

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F = 45N

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5 0
3 years ago
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.

4 0
3 years ago
Read 2 more answers
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