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3 years ago

Why are units of measurement useful?

Physics

1 answer:

Lena [83]3 years ago

3 0

Answer:

Explanation:

because the si system does not need to be completed. also the metric system can also.be used

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(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

3 years ago

What are four factors that can affect the amount of friction between two objects

umka2103 [35]

As the Force of friction is equal to $F _{f} =$ µ $F_{n}$ by using the Law of Action Reaction; as the force normal is the conterforce of gravity

$F _{f} =$ µ $F_{n}$
$F_{f}=$ µ $F_{n}$

Therefore, by looking at the equation we can infer that

The force of friction is directly proportional to the mass, gravitational constant, and the co-efficent of friction

Because the gravitational constant is dependant on gravitation, a planet's mass and radius also affect the force of friction

But DO NOTE:
That the co-effecient of friction is only applicable between two rubbing surfaces and unaffected by gravitational constants.

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Also if this need further explaining, please comment below

7 0

4 years ago

What is the average speed of an object travels 60 meters in 3 seconds?

Solnce55 [7]

Answer:

20 metres per second

Explanation:

60/3

7 0

3 years ago

When the biker is at the top of the ramp, he has a speed of 10 m/s at

oksian1 [2.3K]