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3 years ago

Inertia is the property that every material object has that causes objects to resist changes in its state of motion

2 answers:

4 0

The statement "Inertia is the property that every material object has that causes objects to resist changes in its state of motion" is true

Explanation:

Inertia defines as the resistance produced by the body or an objects to vary in its states of motion, by an external force. An external agent cannot disturb a body, until it has more magnitude than its normal force. Normal force is nothing but due to the mass of the body, being in contact with the ground.

So, inertia is purely dependent on the mass of the body. The unit of inertia is ‘kg’. Inertia, the complete dependent of mass of a body, and also the main reason for the normal force, can stop a body from changing its state.

jonny [76]3 years ago

3 0

True :) good luck on your quiz

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A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th

Iteru [2.4K]

<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

$T^{2}=\frac{4\pi^{2}}{GM}R^{3}$ (1)

Where:

$G$ is the Gravitational Constant

$M=1.9(10)^{27}kg$ is the mass of planet X

$R$ is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

Solving:

$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be 2T.

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3 years ago

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3 years ago

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3 years ago

Ask Your Teacher Suppose the roller coaster below(h1 = 36 m, h2 = 13 m, h3 = 30) passes point A with a speed of 1.00 m/s. If the

Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = ( $\frac{1}{5}$ × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂ = 14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0

3 years ago