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3 years ago

Which of the following is a luminous object? the Moon Jupiter a comet the Sun

Physics

2 answers:

natita [175]3 years ago

5 0

The Sun since it's a big ball of fire

olga nikolaevna [1]3 years ago

4 0

The correct answer is "the sun." Sun is a luminous object. It serves as the source of light that brightens the whole earth. The sun also serves as the power of so that nourishes plants, and even human. It gives light to moon.

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Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

3 years ago

Imagine your teacher asks you to design an experiment where you test the effect of temperature on the growth of a plant. You hav

Dmitry_Shevchenko [17]

Answer:

The correct answer will be-

1. Dependent variable- The growth of plant in the form of height

2. Independent variable- different temperature

3. Constant variable- The amount of water, amount of sunlight, type of soil.

Explanation:

A Scientific experiment must include three types of variables which are: The independent, dependent and the constant variable.

1. Independent variable- The variable which can be modified or changed either on its own or manually. The variable directly influences the variable to be studied. In the given condition, the independent variable is the different temperature provided to the plants.

2. Dependent variable- The variable which is being studied in the experiment and directly influenced by the independent variable is the growth of the plant which is measured in the form of height.

3. Constant variable- The variable which is kept constant throughout the experiment and remains the same which could be the amount of water amount of sunlight and type of soil.

4 0

3 years ago

Mrs. LaCross leaves school and accidentally leaves her coffee mug on the roof of her car as shown in the picture below.

aleksandrvk [35]

Answer:

When she stops at a fast pace the energy and wind will take the cup forward and it will most likeley brake

Explanation:

I'm not entirely sure this is what you were looking for but I hope this helped!

PLEASE MARK ME AS BRAINLIEST

8 0

3 years ago

What order shows increasing frequency for gamma rays, microwaves, visible light, and X-rays?

sergejj [24]

Answer:

Person above me is right, it's B: microwaves, visible light, X-rays, gamma rays

Explanation:

6 0

2 years ago

During an observation, it was noticed that light diffracts as it passes through small slits in a barrier. What does this evidenc

Lynna [10]

Answer:

It reveals that light is a wave

Explanation:

Diffraction is the property of a wave in which there is a bending of the wave about the corners of an obstacle or aperture into the geometrical shadow of the obstacle or aperture.

This simply implies that a wave bends or spreads out when it passes through openings. Since the light diffracts through small slits and diffraction has been shown to occur in water waves and sound waves, this property of diffraction can only be characteristic of a wave and thus, this evidence reveals that light is a wave.

8 0

3 years ago