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3 years ago

How do the units that are used to measure heat differ from the units that are used to measure temperature?

Physics

1 answer:

irakobra [83]3 years ago

6 0

Heat is measured in joules and temperature is measured in F degrees or C degrees.

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To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105

-Dominant- [34]

Explanation:

LD₁ = 10⁵ mm⁻²

LD₂ = 10⁴mm⁻²

V = 1000 mm³

Distance = (LD)(V)

Distance₁ = (10⁵mm⁻²)(1000mm³) = 10×10⁷mm = 10×10⁴m

Distance₂ = (10⁹mm⁻²)(1000mm³) = 1×10¹² mm = 1×10⁹ m

Conversion to miles:

Distance₁ = 10×10⁴ m / 1609m = 62 miles

Distance₂ = 10×10⁹m / 1609 m = 621,504 miles.

7 0

3 years ago

Read 2 more answers

A wave with a low frequency generally has a _____.

Artyom0805 [142]

Its C because if it is a low frequency it will not change much so it will be a longer wavelength and the higher the frequency the shorter the wavelength

8 0

3 years ago

Easy physics question help.!!!

stellarik [79]

Answer: This is not easy lol

Explanation:

3 0

2 years ago

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Un camion de envios se encuentra detenido en una señal de pare, permitiendo que pase una ambulancia. Inicia su recorrido y al ca

Nesterboy [21]

Answer:

$0.741\ \text{m/s}^2$

Explanation:

v = Velocidad final = $40\ \text{km/h}=\dfrac{40}{3.6}\ \text{m/s}$

u = Velocidad inicial = 0

t = Tiempo empleado = 15 s

a = Aceleración

De las ecuaciones cinemáticas tenemos

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{40}{3.6}-0}{15}\\\Rightarrow a=0.741\ \text{m/s}^2$

La aceleración del camión en el primer intervalo de tiempo es $0.741\ \text{m/s}^2$ .

4 0

3 years ago

A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light i

slega [8]

Answer:

The appropriate solution is:

(a) $\frac{1}{4}(I_o)$

(b) $\frac{1}{4} (u_o)$

(c) $\frac{1}{2}B_o$

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

⇒ $\frac{I}{I_o} =\frac{r_o_2}{r_2}$

On applying cross-multiplication, we get

⇒ $I=I_o\times \frac{1_2}{2^2}$

⇒ $=I_o\times \frac{1}{4}$

⇒ $=\frac{1}{4} (I_o)$

(b)

As we know,

⇒ $\frac{u}{u_o} =\frac{I}{I_o}$

By putting the values, we get

⇒ $u=\frac{1}{4}(u_o)$

(c)

⇒ $\frac{B^2}{B_o^2} =\frac{u}{u_o}$

$=\frac{I}{I_o}$

⇒ $B=B_o\times \sqrt{\frac{1}{4} }$

⇒ $=\frac{1}{2}(B_o)$

4 0

3 years ago