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Bingel [31]
3 years ago
11

A set of data is collected for object in an inelastic collision, as recorded in the table.

Physics
1 answer:
TiliK225 [7]3 years ago
5 0

Answer:

To identify the momentum of object 1, you must multiply mass (m) and velocity(v) to find momentum.

Object 1 has momentum of 8 kg. m/s before collision.

Object 1 has momentum of 0 kg. m/s before collision.

The combined mass after the collision had a total momentum of 8 kg. m/s.

Explanation:

Momentum of the object is given by,

Momentum = mass × velocity

For object 1:

Momentum = mass × velocity

Momentum = 2 × 4

Momentum = 8 kg. m/s

For object 2:

Momentum = mass × velocity

Momentum = 6 × 0

Momentum = 0 kg. m/s

For object 1 + object 2:

Momentum = mass × velocity

Momentum = 8 × 1

Momentum = 8 kg. m/s

To identify the momentum of object 1, you must multiply mass (m) and velocity(v) to find momentum.

Object 1 has momentum of 8 kg. m/s before collision.

Object 1 has momentum of 0 kg. m/s before collision.

The combined mass after the collision had a total momentum of 8 kg. m/s.


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4 years ago
The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol
Lesechka [4]

Answer:

\Delta V = 0.053 A

Explanation:

Electric field in a given region is given by equation

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as we know the relation between electric field and potential difference is given as

\Delta V = -\int E. dr

so here we have

\Delta V = - \int (\frac{A}{r^4}) .dr

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here we know that

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so we will have

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3 years ago
9. During an egg toss, a catcher must cushion the egg by maximizing the time it takes to stop the
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Answer:

the impulse experienced by the egg is 0.053 kgm/s.

Explanation:

Given;

mass of the egg, m = 60 g = 0.06 kg

initial velocity of the egg, u = 6 m/s

height moved by the egg, h = 50 cm = 0.5 m

Determine the final velocity of the egg as it moves upward;

v² = u² + 2(-g)h

v² = u² - 2gh

where;

v is the final velocity

-g is negative acceleration due gravity as it moves upward

v² = 6² - 2(9.8 x 0.5)

v² = 26.2

v = √26.2

v = 5.12 m/s

The impulse applied to the egg is the change in linear momentum;

J = ΔP

ΔP = mu - mv

ΔP = m(u - v)

ΔP = 0.06(6 - 5.12)

ΔP = 0.053 kgm/s

Therefore, the impulse experienced by the egg is 0.053 kgm/s.

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