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2 years ago

14

You can feel the heat waves from a small electric heater on your hands and face even though the heater is on the other side of t

he room. Which is this is an example of?

1 answer:

NeX [460]2 years ago

8 0

That's an example of heat transfer by radiation.

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FinnZ [79.3K]

Answer:

A hope this helps

Explanation:

8 0

2 years ago

The process of conversion of solid into gas and again gas into solid is known as

pantera1 [17]

Sublimation is when a solid becomes a gas. And gas to solid is deposition.

6 0

2 years ago

Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu

ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

= 223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

3 0

2 years ago

Una varilla de 5m de longitud y 1.5 cm^2 de sección transversal se alarga 0.10 cm al someterla a una tensión de 700 N. Determina

KengaRu [80]

Bzzhbzaijsb she’s aha ha ha s she’s has shah a sus abuse she d dvhbx

4 0

1 year ago

A displacement is related to time as x = A sin(2?ft), where A and f are constants. Find the dimensions of A. [Hint: A trigonomet

Luba_88 [7]

Answer:

The right option is C. i.e L

Explanation:

Dimension of a quantity is the collection of the indexes of each of the fundamental quantities contained in it.

Therefore, to get the dimension of A in the equation

X=Asin(2(pi)ft)

Making A the subject of the formula

A = X / (sin(2(pi)ft)

Since sin(2(pi)ft which is a trigonometric function does not have a dimension

A = L

The dimension of A is L

5 0

2 years ago