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3 years ago

The unit light-year is a measure of

Physics

2 answers:

erica [24]3 years ago

4 0

The unit light year is a measure of distance.

Virty [35]3 years ago

3 0

One light-year is the distance that light travels in vacuum
in one year. It's a unit of distance.

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A tire has a pressure of 325 kPa at 10°C.

musickatia [10]

5 times that of initial pressure i.e 1625 kpa

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2 years ago

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Can a material have negative permittivity?

sleet_krkn [62]

No it can't it's material

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3 years ago

A crate is pushed horizontally by a horizontal force 527.018 N . Sliding friction resists the motion, and the kinetic coefficien

gulaghasi [49]

Answer:

m= 10 kg a = 52 m / s²

Explanation:

For this problem we must use Newton's second law, let's apply it to each axis

X axis

F - fr = ma

The equation for the force of friction is

-fr = miu N

Axis y

N- W = 0

N = mg

Let's replace and calculate laceration

F - miu (mg) = ma

a = F / m - mi g

a = 527.018 / m - 0.17 9.8

We must know the mass of the body suppose m = 10 kg

a = 527.018 / 10 - 1,666

a = 52 m / s²

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3 years ago

(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 $3.02\times10^{-11}$ .

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2 years ago

Which is the correct order to use in writing the name of an ionic compound?

muminat

Answer:anion followed by cation

Explanation:

8 0

2 years ago

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