The department of insurance and safety is led by what official

Which particles contribute to the net charge and how does each change the net charge?

Licemer1 [7]

The two subatomic particles that contribute to the net charge of an ion are electrons and protons.

Atom is the smallest possible amount of matter which still retains its identity as a chemical element, now known to consist of a nucleus surrounded by electrons.

The atom is made up of three components called subatomic particles as follows;

Protons
Electrons
Neutrons

The proton is the positively charged subatomic particle forming part of the nucleus of an atomwhile the electron is the subatomic particle having a negative charge and orbiting the nucleus.

This suggests that the two subatomic particles that contribute to the net charge of an ion are electrons and protons. That is;

Net charge = protons - electrons

Learn more about subatomic particles at:brainly.com/question/13303285

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3 0

1 year ago

The value of a scientific variable ____. A) can change

gregori [183]

The answer is A: can change

5 0

2 years ago

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8. While taking a measurement, Ajay put the 2nd mark of the scale to the edge of the line and the mark that pointed to the end o

Leni [432]

Answer:

The length of line is 78 cm or 0.78 m.

Explanation:

initial reading 2 mark

final reading 80 cm

The length of the line

= final reading - initial reading

= 80 - 2

= 78 cm

1 cm = 0.01 m

So, 78 cm = 0.78 m

4 0

1 year ago

Which of the following is an example of balanced forces?

Sergio [31]

A seesaw remains stationary when two students of equal weight sit on the ends

c

6 0

2 years ago

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Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$