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SIZIF [17.4K]
3 years ago
15

If a car that is moving 20.0 m/s has a momentum of 29000 kg·m/s, what mass is the car?

Physics
1 answer:
Strike441 [17]3 years ago
3 0

Answer:

Mass = 1450kg

Explanation:

P = M * V (where p is momentum, m is mass and v is velocity)

29000 = 20 * M

M = 29000 / 20

M= 1450 kg

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Much energy as would Microraptor gui have to expend to fly with a speed of 10 m/s for 1.0 minutes is 486 J.

The first step is to find the energy that Microraptor must release to fly at 10 m/s for 1.0 minutes. The energy that Microraptor must expend to fly can be found using the relationship between Power and Energy.

P = E/t

Where:

P = power (W)

T = time (s)

Now, a minimum of 8.1 W is required to fly at 10 m/s. So, the energy expended in 1 minute (60 seconds) is

P = E/t

E = P x t

E = 8.1 x 60

E = 486 Joules

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A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.
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The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>
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  • Distance moved by the box, d = 2.5 m
  • Coefficient of friction, = 0.35
  • Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>
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The work done on the box by the applied force is calculated as;

W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

W = (0) d \times cos(30)\\\\W = 0 \ J

The work done by the force of gravity is calculated as follows;

W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J

The work done on the box by the normal force is calculated as follows;

W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J

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8 0
2 years ago
How will the frequency of a wave appear to change if the source of the wave is moving toward an observer
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pitch goes up on approach ... Doppler effect

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my hypothesis is that If you drop a piece of buttered toast, it will land butter side down.


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Now differentiating the equation (4) w.r.t. 'b', we get

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This is the amount of energy stored in the conductor.

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