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djyliett [7]
3 years ago
12

How do meteorites differ from meteors and meteoroids?

Physics
2 answers:
Evgesh-ka [11]3 years ago
8 0

Answer:

When meteoroids enter Earth's atmosphere (or that of another planet, like Mars) at high speed and burn up, the fireballs or “shooting stars” are called meteors. When a meteoroid survives a trip through the atmosphere and hits the ground, it's called a meteorite.

Explanation:

Ira Lisetskai [31]3 years ago
6 0

Answer: Meteoroid: Small particle from a comet or asteroid orbiting the Sun. Meteor: The light phenomena which results from a meteoroid entering the Earth's atmosphere and vaporizing; this is basically a shooting star. Meteorite: A meteoroid that survives through the Earth's atmosphere and lands on the Earth's surface.

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Moving a magnet inside a coil of wire will induce a voltage in the coil. How can the voltage in the coil be increased?
mr_godi [17]
As the magnet is moved inside a coil of wire, the number of lines of magnetic field passing through the coil changes. Faraday stated that : it is the change in the number of field lines passing through the the coil of wire that induces emf in the loop. Specifically, it is the rate of change in the number of magnetic field lines passing through the loop that determines the induced emf. There is a term called magnetic flux same as electric flux, this magnetic flux can be a measure of the number of field lines passing through a surface. It is given by ( Φ=ΣB. dA. Where B is magnetic field and dA is small elementary area). The induced emf is given by (ξ = dΦ/dt). This equation states that THE MAGNITUDE OF THE INDUCED CURRENT IN A CIRCUIT IS EQUAL TO THE RATE AT WHICH THE MAGNETIC FLUX THROUGH THE CIRCUIT IS CHANGING WITH TIME. So more rapid you move the coil, more will be the change in flux and hence more emf will be produced. So option D is the correct answer. I hope this long description will help you out.
5 0
3 years ago
Read 2 more answers
"3.) By Newton’s 3rd Law, a train pulls backward on its engine exactly as hard as the engine pulls forward on the train. Since N
dem82 [27]

Answer:

The force is applicable according to the newton's third law of motion but the force on the engine is compensated in the form of stess on the fixed parts and rigid links whereas the wheel is free to roll.

Explanation:

This interpretation that the engine applies a force on the train and the train also applies an equal force on the engine and hence the train should not move is wrong because the engine imparts a rotational force in the form of torque on the train and the train imparts an equal force on the engine in the opposite direction but the engine is fixed like a structure on the chassis of the train and consists of rigid links which resist the motion and deformation as compared to the relative motion between the wheels and the rail tracks.

3 0
3 years ago
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad / s). if a particular disk is spun at 79
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3 years ago
With what speed must m1 rotate in a circle of radius r if m2 is to remain hanging at rest?
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5 0
4 years ago
A 3 kg block is released from rest to slide down a ramp with friction. The length that the block slides is 2 meters and the angl
Arte-miy333 [17]

Answer:

(a). The acceleration is 3.20 m/s².

(b). The amount of work done by each force is 19.2 J.

(c). The total work done on the block is 19.2 J.

(d). The final speed of the block is 6.26 m/s.

Explanation:

Given that,

Mass of block = 3 kg

Distance = 2 m

Angle = 30°

Coefficient of kinetic friction = 0.20

(a). We need to calculate the acceleration

Using balance equation of force

N = mg\co\theta

mg\sin\theta-f_{k}=ma

a = g\sin\theta-\mu g\cos30

Put the value into the formula

a=g(\sin30-0.20\cos30)

a=9.8(\dfrac{1}{2}-0.20\times\dfrac{\sqrt{3}}{2})

a=3.20\ m/s^2

The acceleration is 3.20 m/s².

(b). We need to calculate the amount of work done by each force

Using formula of work done

Normal force is

N = mg\cos30

So due to this the net force is zero then the no work done by reaction force.

By another force,

W= F\times d

W=ma\times d

Put the value into the formula

W= 3\times3.20\times2

W=19.2\ J

The amount of work done by each force is 19.2 J.

(c). We need to calculate the total work done on the block

The total work done on the block is 19.2 J.

(d). We need to calculate the final speed of the block

Using equation of motion

v^2=u^2+2gs

Put the value into the formula

v^2=0+2\times9.8\times2

v=\sqrt{2\times9.8\times2}

v=6.26\ m/s

The final speed of the block is 6.26 m/s.

Hence, This is the required solution.

4 0
3 years ago
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