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3 years ago

A train of mass 3.3 × 10^6 kg is moving at a constant speed up a slope inclined at an angle of 0.64°

Physics

1 answer:

AnnyKZ [126]3 years ago

4 0

Answer:

C. 39 m/s

Explanation:

First we need to calculate the total force required to move the train along the inclined plane. So, it is clear that the work done will be equal to the component of the weight that is parallel to the inclined plane, because there is no frictional force present:

Force = F = mg Sin θ

where,

m = mass of train = 3.3 x 10⁶ kg

g = 9.8 m/s²

θ = Angle of Inclination = 0.64°

Therefore,

F = (3.3 x 10⁶ kg)(9.8 m/s²)Sin 0.64°

F = 3.612 x 10⁵ N

Now, the formula for power is:

P = FV

V = P/F

where,

V = Velocity of Train = ?

P = Power of Engine = 14 MW = 1.4 x 10⁷ W

Therefore,

V = 1.4 x 10⁷ W/3.612 x 10⁵ N

V = 38.75 m/s

which is approximately equal to:

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Given:

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Thus a=- 9.8 m/s^2

Now consider the equation

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Where u is the initial velocity which is measured in m/s

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s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.

Substituting the given values in the above formula we get

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3 years ago

Two electromagnets contain the same types of magnets

Ira Lisetskai [31]

Answer: The first electromagnet has a more powerful current than

the second

Explanation:

Since the two electromagnets contain the same types of magnets and wires. If the magnet In the first moves much faster than the second. Therefore:

The first electromagnet has a more powerful current than the second

Because the induced EMF is proportional to the induced current.

Where the induced EMF depends on the speed of the magnet according to the formula below

EMF = BVL

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3 years ago

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

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3 years ago