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EleoNora [17]
2 years ago
11

A 3.0-kg ball with an initial velocity of (4.0i + 3.0j) m/s collides with a wall and rebounds with a

Physics
1 answer:
irga5000 [103]2 years ago
3 0

The impulse exerted on the ball by the wall is 24i kgm/s.

<h3>Impulse exerted by the ball on the wall</h3>

The impulse exerted by the ball on the wall is the change in the linear momentum of the ball.

J = ΔP

ΔP = Pf - Pi

P = mv

where;

  • m is mass of the ball
  • v is the velocity

ΔP = 3(4.0i + 3.0j) - 3(-4.0i + 3.0j)

ΔP = (12i + 9j) - (-12i - 9j)

ΔP = 24i kgm/s

Thus, the impulse exerted on the ball by the wall is 24i kgm/s.

Learn more about impulse here: brainly.com/question/904448

#SPJ1

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5 0
2 years ago
Urgently!
adelina 88 [10]

Answer:

0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli

1.1m to 0.04125 m

Explanation:

T= 1/f=

if f= 300Hz then T = 1/300 =0.003333 s

if f= 8000 then T= 1/8000 = 0.000125s

now v=f×wave length

or wavelength = speed/ frequency

when f = 300 Hz

wavelength = 330/300=1.1 m

wavelength = 330/8000 = 0.04125m

note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s

6 0
3 years ago
A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As th
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Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

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\Delta L=74.7-58.6

\Delta L=16.1\ m

For an open-open tube,

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Using formula of wavelength

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\lambda=32.2\ cm

We need to calculate the frequency of the tuning

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f=\dfrac{v}{\lambda}

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f=\dfrac{343}{32.2\times10^{-2}}

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