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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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Water flowing through a cylindrical pipe suddenly comes to a section of pipe where the diameter decreases to 86% of its previous

Orlov [11]

Answer:

Explanation:

The speed of the water in the large section of the pipe is not stated

so i will assume 36m/s

(if its not the said speed, input the figure of your speed and you get it right)

Continuity equation is applicable for ideal, incompressible liquids

Q the flux of water that is Av with A the cross section area and v the velocity,

so,

$A_1V_1=A_2V_2$

$A_{1}=\frac{\pi}{4}d_{1}^{2} \\\\ A_{2}=\frac{\pi}{4}d_{2}^{2}$

the diameter decreases 86% so

$d_2 = 0.86d_1$

$v_{2}=\frac{\frac{\pi}{4}d_{1}^{2}v_{1}}{\frac{\pi}{4}d_{2}^{2}}\\\\=\frac{\cancel{\frac{\pi}{4}d_{1}^{2}}v_{1}}{\cancel{\frac{\pi}{4}}(0.86\cancel{d_{1}})^{2}}\\\\\approx1.35v_{1} \\\\v_{2}\approx(1.35)(38)\\\\\approx48.6\,\frac{m}{s}$

Thus, speed in smaller section is 48.6 m/s

3 0

3 years ago

A wave with a high frequency generally has a _____.

il63 [147K]

High frequency = D, short wavelength

4 0

3 years ago

Type the correct answer in the box. Round your answer to the nearest tenth. Dina has a mass of 50 kilograms and is waiting at th

katen-ka-za [31]

I guess once you input the numbers into the correct places into the equation it would look like this :
PE = 50 * 9.8 * 5.0
PE = 2,450

6 0

3 years ago

Read 2 more answers

1 oz = _____________ mg

Luba_88 [7]

Answer:

28349.5 Mg

Explanation:

4 0

3 years ago

A spotlight on the ground shines on a wall 12m away. If a man 2m tall walks from the spotlight toward a building at a speed of 1

exis [7]

we suppose that x be distance of man from spot light is
so 12-x is distance from man to wall
we will draw first triangle ABC, where
a= spotlight (on ground)
b= man (feet)
c= man (head)
we know that AB = x and BC = 2
by extend line AB to the wall at point D and extend line AC to the wall at point E
AD = 12 (distance from spotlight to wall)
DE = s (length of shadow)
Now triangles ABC and ADE are similar
Therefore DE/AD = BC/AB
s / 12 = 2 / x
s = 24 / x
Differentiate both sides with respect to t
ds/dt = -24/x² dx/dt
Man is walking toward building at speed of 16 ms
dx/dt = 16
Man is 4 m from the building/wall
12 - x = 4
x = 8
Find ds/dt when x = 8 and dx/dt = 16
ds/dt = -24/x² dx/dt
ds/dt = -24/(64) * 16
ds/dt = -6

So length of shadow is decreasing at rate of 6 m/s
hope it helps

4 0

3 years ago