Find f (the force) in relation to moments.

Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

VARVARA [1.3K]

The rate of heat loss by radiation is equal to -207.5kW

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0

3 years ago

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of

Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

$E = \frac{mg}{q}$

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

$q = \frac{mg}{E}$

Replacing,

$q = \frac{(3.37*10^{-9})(9.8)}{11000}$

$q = 3.002*10^{-12}C$

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

$q = ne$

Here,

n = Number of electrons

e = Charge of each electron

$n = \frac{q}{e}$

Replacing,

$n = \frac{3.002*10^{-12}}{1.6*10^{-19}}$

$n = 2.44*10^7$

Therefore the number of electrons that reside on the drop is $2.44*10^7$

5 0

3 years ago

Dont know what to type here ok chat meh

never [62]

ogckcigxxkgxkgxkgxixogxigxixgxgifkcgo

5 0

2 years ago

What is the relative size and composition of the universe, a galaxy, and a solar system? Is the universe endless?

vichka [17]

Yas, the universe is endless/infinte

7 0

2 years ago

Read 2 more answers

An object is dropped from a vertical distance of 25.5 m above the ground, and it takes 2.28 sec to fall that distance. A second

DENIUS [597]

Answer:

The second object takes 2.28 s to fall the 25.5 m.

Explanation:

In this case, both objects take the same time to fall, since no vertical velocity is added to any of them.

You can also confirm this by sepparating the second's object movement into its two directions: in the horizontal one, we have linear uniform motion, and in the vertical one, we have free fall, with exactly the same characteristics as for the first object.

4 0

3 years ago

Find f (the force) in relation to moments.​

Find f (the force) in relation to moments.