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Vsevolod [243]
3 years ago
9

How many joules are in 10,000 Electron Volts

Physics
2 answers:
tatiyna3 years ago
7 0
1.602177e-15 joules are present in 10,000 Electron Volts :)
Semenov [28]3 years ago
5 0
1.602177e-15 joules are in 10,000 electron volts

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If
timama [110]

Answer:

Workdone = 1960 Joules.

Explanation:

Given the following data;

Mass = 5kg

Force = 49N

Height (distance) = 40m

To find the workdone;

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 49*40

Workdone = 1960 Joules.

Therefore, the amount of work done on the bowling ball to lift it is 1960 Joules.

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3 years ago
In the nucleus, the a)...... have b)...... charge.
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C:carry 
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3 years ago
A 900 N student runs up the stairs 3.5 m high in 12<br> seconds. How much POWER do they generate?
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Answer:

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4 0
3 years ago
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
When an object of weight w is suspended from the center of a massless string?
igomit [66]
I attached a picture of the diagram associated with this question.

Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ

The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ

4 0
3 years ago
Read 2 more answers
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