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Sonbull [250]
2 years ago
11

A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .if after the collision it moves with a velocity of 2

.0m/s in the opposite direction,calculate the change in momentum
Physics
1 answer:
ch4aika [34]2 years ago
7 0

Answer:

-1.2 kg - m/s

Explanation:

\pink{\frak{Given}}\begin{cases}\textsf{ A body of mass 100g moving with a velocity of 10.0m/s collides with a wall .}\\\textsf{ After the collision it moves with a velocity of 2.0m/s in the opposite direction.}\end{cases}

And we need to find out the change in momentum of the body . Here ,

  • velocity before collision (u) = 10m/s
  • velocity after collision (v) = 2m/s .

We know that momentum is defined as amount of motion contained in a body . Mathematically ,

\sf\longrightarrow momentum (p)= mass(m) * velocity(v)

Therefore change in momentum will be,

\sf\longrightarrow \triangle p = mv - mu

Since the direction of velocity changes after the collision , the velocity will be -2m/s .

\sf\longrightarrow \Delta p = 100g( -2m/s -10m/s) \\

\sf\longrightarrow \Delta p =\dfrac{100}{1000}kg ( -12m/s)  \\

\sf\longrightarrow \Delta p   = 0.1 kg * -12m/s \\

\sf\longrightarrow \boxed{\bf \Delta p = -1.2 \ kg-m/s} \\

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Answer:

1. The image is real

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Explanation:

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1/5.01 = 1/13.7 + 1/q

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Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,

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q = 49.3 cm behind the second lens

Using that information, since q is positive, the image is real

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The height can be found from the two magnifications.

m = -q/p

m1 = -7.9/13.7 = -.577

m2 = -49.3/54.6 = -.903

Net m = (-.577)(-.903) = .521

Then, m = h'/h

.521 = h'/5.85

3. h' = 3.05 mm

4. For the fourth answer, since the overall magnification is positive, the final image is upright

5 0
3 years ago
How do you calculate the magnitude of a force?
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3 0
3 years ago
In an experiment, the 1 kg cart collides with a 3 kg cart but doesn’t stick to it. Instead, the 3 kg cart gets knock forward by
notka56 [123]

Explanation:

Given that,

Mass of the cart, m_1=1\ kg

Mass of the cart 2, m_2=3\ kg            

Final speed of cart 2, v_2=0.3\ m/s

Final speed of cart 1 is 0 as it comes to rest.

Let us assume that the initial velocity of the cart 2 is 0. So using the conservation of linear momentum as :

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A roller-skate with a mass of 10kg has a force of 12N exerted on it. At what rate did the skate accelerate?
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