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Alenkinab [10]
3 years ago
13

A 4.3-g object moving to the right at 22 cm/s makes an elastic head-on collision with a 8.6-g object that is initially at rest.

what is the velocity of the 4.3-g mass after the collision?
Physics
1 answer:
3241004551 [841]3 years ago
7 0
The right<span> at +20.0 </span>cm/s makes<span> an </span>elastic head<span>-on </span>collision<span> with a 10.0 </span>g object<span> that </span>makes<span> an</span>elastic head<span>-on </span>collision<span> with a 10.0 </span>g object<span> that is </span>initially<span> at </span>rest<span>.(b) Find the fraction of the </span>initial<span>kinetic energy transferred to the 10.0 </span>g object<span>.of small </span>mass<span> before and </span>after collision; V=velocity<span> of big </span>mass after collision<span>.</span>
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A student tested a variety of interacting forces to determine how they would result in motion of an object. If the student used
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Answer:

D) The variable shown by letter C would result in a movement of the object to the right.

Explanation:

7 0
3 years ago
I would like to know the correct answer for this
Mnenie [13.5K]

The last choice on the list is the correct one, for both #2 and #3.

3 0
4 years ago
A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po
Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

      gh1 + 0.5V1^{2} = gh2

      V1 = \sqrt{\frac{gh2 - gh1}{0.5}}

now we can substitute all required values into the equation above.

V1 = \sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}

V1 = \sqrt{\frac{32.34}{0.5}}

V1 =8.1 m/s

8 0
4 years ago
If a 25kg mass is dropped from a height of 5 meters what’s the work done by gravity
Elanso [62]

Answer:

Explanation:

M = 25kg

h = 5m

g = 9.8m/s2

Work done = mgh

Work done = 25 x 9.8 x 5

Work done = 1225J

8 0
3 years ago
If an ocean wave passes a stationary point every 5 s and has a velocity of 10 m/s, what is the wavelength of the wave
Rainbow [258]

Answer:

As Per Provided Information

Velocity of wave v is 10m/s

These ocean wave passes a stationary point every 5 s ( It's time period)

First we calculate the frequency of ocean wave .

<u>Using</u><u> Formulae</u>

\blue{\boxed{\bf \:  \nu =  \cfrac{1}{v}}}

here

v is the velocity of wave .

\sf\longrightarrow \nu \:  =  \cfrac{1}{10}  \\  \\ \\  \sf\longrightarrow \nu \:  = 0.1Hz

Now calculating the wavelength of the wave .

<u>Using </u><u>Formulae </u>

\boxed{ \bf \lambda =  \cfrac{v}{ \nu}}

Substituting the value and we obtain

\sf \longrightarrow \lambda \:  =  \cfrac{10}{0.1}  \\  \\  \\ \sf \longrightarrow \lambda \:  =   \cancel\cfrac{10}{0.1}  \\  \\  \\ \sf \longrightarrow \lambda \:  =100m

<u>Therefore</u><u>,</u>

  • <u>Wavelength </u><u>of </u><u>the </u><u>wave </u><u>is </u><u>100 </u><u>metres</u><u>.</u>
8 0
3 years ago
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