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Soloha48 [4]
2 years ago
9

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

Physics
1 answer:
Masja [62]2 years ago
7 0

Answer:

B

Explanation:

i7tigp79ug978uyioh;

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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 <u>-207.5kW</u>

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<br><br><br><br>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 <em>no vertical velocity is added </em>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 <em>linear uniform motion, </em>and in the vertical one, we have <em>free fall, </em>with exactly the same characteristics as for the first object.

4 0
3 years ago
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