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alina1380 [7]
3 years ago
5

Relate speed, velocity, and acceleration to objects in motion.

Physics
1 answer:
stiks02 [169]3 years ago
6 0

Answer:

speed- distance traveled per unit of time.

velocity-rate of change of its position

acceleration-rate of change of the velocity

Explanation:

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What did Rutherford’s model of the atom include that Thomson’s model did not have?
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If the index of refraction of a medium is 1.4, determine the speed of light in that medium.
Damm [24]

The speed of light in that medium is 2.14 \times 10^8 \ m/s.

<u>Explanation:</u>

It is known that the light's speed is constant when it travels in vacuum and the value is 3 \times 108 m/s. When the light enters another medium other than vacuum, its speed get decreased as the light gets refracted by an angle.

The amount of refraction can be determined by the index of refraction or refractive index of the medium. The refraction index is measured as the ratios of speed of light in vacuum to that in the medium. It is represented as  η = \frac {c}{v}

So, here η is the index of refraction of a medium which is given as 1.4, c is the light's speed in vacuum (3 \times 10^8 ms^-^1) and v is the light's speed in that medium which we need to find.

1.4=  \frac{(3 \times 10 ^ 8)} {v}

v=  \frac {(3 \times 10^8)}{1.4} =2.14 \times 10^8 \ m/s

Thus the speed of light in that medium is 2.14 \times 10^8 \ m/s.

3 0
3 years ago
A pendulum oscillates with a period T. If both the mass of the bob and the length of the pendulum are doubled, the new period wi
Snezhnost [94]

Answer:

T\sqrt{2}

Explanation:

First let us write out the formula connecting the simple parameters in a system of simple pendulum

\\ T=2\pi \sqrt{\frac{L}{g} }\\

Where

T= period of oscillation,

L=length of the pendulum

g= acceleration due to gravity

from the above formula, it is obvious that Mass has no effect on the period(T).

Again let see the relationship that exist between the period(T) and the length(L).

let assume 2\pi \frac{1}{\sqrt{g}} = k , then we can have

T= k\sqrt{L}\\

We can conclude that the period is directly proportional to the square-root of the length.

we can vary the constant K to have

\frac{T_{1} }{\sqrt{L_{1}}}=\frac{T_{2} }{\sqrt{L_{2}}} =...=\frac{T_{n} }{\sqrt{L_{n}}}

Now back to the question, if the length was  doubled, we have the

L_{2}=2L_{1}

using the varying equation, we can substitute values and arrive at

\frac{T_{1} }{\sqrt{L_{1}}}=\frac{T_{2} }{\sqrt{2L_{1}}} \\

carrying out simple operation, we arrive at

\frac{T_{1}*\sqrt{2}*\sqrt{L_{1} }}{\sqrt{L1} }=T_{2}\\

Finally we have

T_{2}=T_{1} \sqrt{2} \\

hence we can conclude that the When the length is doubled, the period is increase by a factor of root 2

4 0
3 years ago
Grade Sem 1
Anvisha [2.4K]

Answer:

anything that occupies space and has mass called matter

3 0
3 years ago
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