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3 years ago

What type of heat does not require matter?

Physics

2 answers:

koban [17]3 years ago

8 0

Radiation is the only method of heat transfer that does not require matter.

Facts about Radiation :

The energy from the sun is received in the form of radiation.
It travels in the form of waves.
It can be absorbed by substances in its path.
Heat Radiation travels through air.
Black surfaces are the emitters of heat radiation.
Shiny surfaces are the reflectors of heat radiation.
Heat transfer by Radiation doesn't require any medium.

Lana71 [14]3 years ago

4 0

It would be Thermal Radiation

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otez555 [7]

Answer: 3.1158

Explanation: THERE

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2 years ago

Which statement best describes a characteristic of gases?

salantis [7]

Assumes the shape and volume of its container
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3 years ago

A 20-turn coil of area 0.32 m2 is placed in a uniform magnetic field of 0.055 T so that the perpendicular to the plane of the co

makvit [3.9K]

Answer:

1.5 * 10^-2 Tm^2

Explanation:

Electric Flux = B.A cos(theta)

B = 0.055 T

A = 0.32 m^2

theta = 30

Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2

5 0

3 years ago

A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W

dezoksy [38]

Answer:

Explanation:

a ) Slit separation d = .1 x 10⁻³ m

Screen distance D = 4 m

wave length of light λ = 650 x 10⁻⁹ m

Width of central fringe = λ D / d

= $\frac{650\times10^{-9}\times4}{.1\times10^{-3}}$

= 26 mm

b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm

c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to

λ / d

= $\frac{650\times10^{-9}}{.1\times10^{-3}}$

= 6.5 x 10⁻³ radian.

8 0

3 years ago

If the sprinter accelerates at that rate for a distance of 15 m, and then maintains the velocity he has at that point for the re

ahrayia [7]

Answer:

The time for the entire race is 11.39 sec.

Explanation:

Given that,

Distance = 15 m

Remainder distance = 100 m

Suppose A sprinter begins a race with an acceleration of 3.4 m/s².

We need to calculate the time

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value in the equation

$15=0+\dfrac{1}{2}\times3.4\times t^2$

$t^2=\dfrac{30}{3.4}$

$t=2.97\ sec$

We need to calculate the final velocity of sprinter

Using equation of motion again

$v=u+at$

Put the value into the formula

$v=0+3.4\times2.97$

$v=10.09\ m/s$

We need to calculate the distance covers by sprinter

$d=100-15=85\ m$

The sprinter need to covers only 85 m.

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t'=\dfrac{85}{10.09}$

$t'=8.42\ sec$

We need to calculate the time for the entire race

$t''=t+t'$

Put the value into the formula

$t''=2.97+8.42$

$t''=11.39\ sec$

Hence, The time for the entire race is 11.39 sec.

4 0

3 years ago