Two bodies, one hot and the other cold kept in vacuum.what will happen to the tempreture of bodies after some time.

Answers

Answer 1

Answer: Hot body will lose heat from it, and that heat will goes out from it through radiation, so it's temperature will decrease after some time.

In same manner, cold body will take the heat, and it's temperature will increase

Hope this helps!

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A rider on a bike with the combined mass of 100 kg attains a terminal speed of 15 m/s on a 12% slope. Assuming that the only forces affecting the speed are the weight and the drag, calculate the drag coefficient. The frontal area is 0.9 m2.

Answers

Answer:

7.84

Explanation:

Draw free body diagram and put all forces on it. Forces are

Gravity in direction of slope = $(12)/(100) *100*12$ =117.6newton
Viscous force(opposite to slope) after attaining terminal speed =k*V=k*15

As the bike+man system has attained a terminal velocity thus acceleration is zero .

Both forces are opposite then equate them

117.6=k.15

k=7.84

Here k is drag coefficient.

Answer:

0·95

Explanation:

Given the combined mass of the rider and the bike = 100 kg

Percent slope = 12%

∴ Slope = 0·12

Terminal speed = 15 m/s

Frontal area = 0·9 m²

Let the slope angle be β

tanβ = 0·12

As it attains the terminal speed, the forces acting on the combined rider and the bike must be balanced and therefore the rider must be moving download as the directions of one of the component of weight and drag force will be in opposite directions

The other component of weight will get balance by the normal reaction and you can see the figure which is in the file attached

From the diagram m × g × sinβ = drag force

Drag force = 0·5 × d × $C_(D)$ × v² × A

where d is the density of the fluid through which it flows

$C_(D)$ is the drag coefficient

v is the speed of the object relative to the fluid

A is the cross sectional area

As tanβ = 0·12

∴ sinβ = 0·119

Let the fluid in this case be air and density of air d = 1·21 kg/m³

m × g × sinβ = 0·5 × d × $C_(D)$ × v² × A

100 × 9·8 ×0·119 = 0·5 × 1·21 × $C_(D)$ × 15² × 0·9

∴ $C_(D)$ ≈ 0·95

∴ Drag coefficient is approximately 0·95

A concrete highway is built of slabs 18.0 m long (at 25 °C). How wide should the expansion cracks be (at 25 °C) between the slabs to prevent buckling if the annual extreme temperatures are −32 °C and 52 °C?(the coefficient of linear expansion of concrete is 1.20 × 10 − 5 °C-1) g

Answers

To solve the problem it is necessary to apply the concepts related to thermal expansion of solids. Thermodynamically the expansion is given by

$\Delta L = L_0 \alpha \Delta T$

Where,

$L_0 =$ Original Length of the bar

$\Delta T$ = Change in temperature

$\alpha$ = Coefficient of thermal expansion

On the other hand our values are given as,

$L_0 = 18m$

$\alpha = 12*10^(-6)/\°C$

$T_2 = 52\°C$

$T_1= 25\°C$

Replacing we have,

$\Delta L = L_0 \alpha (T_2-T_1)$

$\Delta L = (18)(12*10^(-6))(52-25)$

$\Delta L = 5.832*10^(-3)m$

The width of the expansion of the cracks between the slabs is 0.5832cm

The width of the expansion cracks between the slabs to prevent buckling should be 0.5832cm.

How to calculate width?

According to this question, the following information are given:

Lo = Original length of the bar
∆T = Change in temperature
α = Coefficient of thermal energy

The values are given as follows:

Lo = 18m
T1 = 25°C, T2 = 52°C
α = 12 × 10-⁶/°C

∆L = Loα (T2 - T1)

∆L = 18 × 12 × 10-⁶ (27)

∆L = 3.24 × 10-⁴ × 18

∆L = 5.832 × 10-³m

Therefore, the width of the expansion of the cracks between the slabs is 0.5832cm.

Learn more about width at: brainly.com/question/26168065

Ken Griffey, Jr's warehouse shot in the 1933 home run derby travelled 93 feet per second for 5 seconds. How far did he hit the ball?

Answers

Answer:

465 feet because 93*5 = 465, btw that was 1993 not 1933

Explanation:

A machine can make doing work easier by reducing the force exerted, changing the distance over which the force is exerted, or changing the direction of the force.True OR False

HELP ME!!!!!!¡!!!!

Answers

I believe the correct answer is true. A machine can make doing work easier by reducing the force exerted, changing the distance over which the force is exerted, or changing the direction of the force. Hope this answer the question.

Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

Answers

Answer:

a) The UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

b) The radiation with a frequency of $7.9x10^(14)Hz$ belong to the UV-A category.

Explanation:

(a) Find the range of frequencies for UV-B radiation.

Ultraviolet light belongs to the electromagnetic spectrum, which distributes radiation along it in order of different frequencies or wavelengths.

Higher frequencies:

Gamma ray

X ray

Ultraviolet rays

Visible region

Lower frequencies:

Infrared

Microwave

Radio waves

That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it. Any of those radiations will have a speed of $3x10^{8]m/s$ in vacuum.

The velocity of a wave can be determined by means of the following equation:

$c = \nu \cdot \lambda$ (1)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength.

Then, from equation 1 the frequency can be isolated.

$\nu = (c)/(\lambda)$ (2)

Before using equation 2 to determine the range of UV-B it is necessary to express $\lambda$ in units of meters in order to match with the units from c.

$\lambda = 320nm . (1m)/(1x10^(9)nm)$ ⇒ $3.2x10^(-7)m$