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2 years ago

For your answer to this problem, just type in the numerical magnitude of the momentum - no units.

Physics

1 answer:

stepan [7]2 years ago

3 0

Answer:

120 kg•m/s.

Explanation:

From the question given above, the following data were obtained:

Case 1

Mass of object = M

Velocity of object = V

Momentum = 15 kg•m/s

Case 2

Mass of object = 2M

Velocity of object = 4V

Momentum = ?

Momentum is defined as follow:

Momentum = mass × velocity

The momentum of object in case 2 can be obtained as follow:

From case 1

Momentum = mass × velocity

15 = M × V

15 = MV ....... (1)

From case 2:

Momentum = mass × velocity

Momentum = 2M × 4V

Momentum = 8MV ....... (2)

Finally , substitute the value of MV in equation 1 into equation 2.

Momentum = 8MV

MV = 15

Momentum = 8 × 15

Momentum = 120 kg•m/s

Therefore, an object with a mass of 2M and 4V would have a momentum of 120 kg•m/s

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A lightning bolt with 13 kA strikes an object for 14 μ s. How much charge is deposited on the object?

pogonyaev

Answer:

0.182C

Explanation:

Using Q= It

= 13x10^3 . 14x10^-6

= 0.182C

8 0

3 years ago

An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a

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Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = $3.1$ m

Width of the window = $2.1$ m

Area of the window = $(3.1\times 2.1) = 6.51\ m^2$

Difference in air pressure = Inside pressure - Outside pressure

= $(1.0-0.954)$ atm = $0.046$ atm

Conversion of the pressure in its SI unit.

⇒ $1$ atm = $101325$ Pa

⇒ $0.046$ atm = $0.046\times 101325 =4660.95$ Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ $Pressure=\frac{Force }{Area}$

⇒ $Force =Pressure\times Area$

⇒ Plugging the values.

⇒ $Force =4660.95\times 6.51$

⇒ $Force=30342.78$ Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

3 0

3 years ago

A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo

Charra [1.4K]

Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, $\omega=0.47\ rev/s=2.95\ rad/s$

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

$I={m_1r^2}$

$I={52\times 3.9^2}$

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

$I_2=\dfrac{m_2r^2}{2}$

$I_2=\dfrac{118\times (3.9)^2}{2}$

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

$I=I_1+I_2$

$I=790.92+897.39$

I = 1688.31 kg-m²

The angular momentum of the system is :

$L=I\times \omega$

$L=1688.31 \times 2.95$

$L=4980.5\ kg-m^2/s$

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Answer:

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Explanation:

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Explanation:

Below is an attachment containing the solution.

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