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3 years ago

What is the primary visible color of an emission nebula?

1 answer:

7 0

In emission nebulae, there are interstellar clouds of hydrogen, which glow red because of the intense radiation of hot stars inside the nebula

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A long straight wire has fixed negative charge with a linear charge density of magnitude 4.6 nC/m. The wire is to be enclosed by

Semmy [17]

Answer: -33.3 * 10^9 C/m^2( nC/m^2)

Explanation: In order to solve this problem we have to use the gaussian law, the we have:

Eoutside =0 so teh Q inside==

the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.

Then we simplify and get

σ= -4.6/(2*π*b)= -33.3 nC/m^2

7 0

3 years ago

Which form of energy is equal to the sum of an object’s kinetic and potential energy?

Natali5045456 [20]

Answer:

Mechanical Energy

Explanation:

The sum of kinetic energy and potential energy of an object is its total mechanical energy.

4 0

3 years ago

write an essay describing and predicting the effects of fitness-related stress management techniques on the body.

Tatiana [17]

This is something u are going to have to do

7 0

3 years ago

A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric fiel

nexus9112 [7]

Answer:

The electric flux is $280\ \rm N.m^2/C$

Explanation:

Given:

Radius of the disc R=0.50 m
Angle made by disk with the horizontal $\theta=30^\circ$
Magnitude of the electric Field $E=713.0\ \rm N/C$

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

$\phi=\int E.dA$

where

$\phi$ is the total Electric Flux
E is the Electric Field
dA is the Area through which the electric flux is to be calculated.

Now according to question we have

$=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C$

Hence the electric flux is calculated.

8 0

3 years ago

The half-life of a certain radioactive substance is 12 hours. There are 19 grams present initially. a. Express the amount of su

neonofarm [45]

Answer:

(a) $N=19\times e^{-\lambda t}$

(b) 15 hours

Explanation:

half life, T = 12 hours

No = 19 g

(a) Let N be the amount remaining after time t.

Let λ be the decay constant.

$\lambda =\frac {0.6931}{T}$

The equation of radioactivity used here is given by

$N=N_{o}e^{-\lambda t}$

$N=19\times e^{-\lambda t}$

(b) N = 8 gram

Substitute the values in above equation

$\lambda =\frac {0.6931}{12}$

λ = 0.0577 per hour

So, $8=19\times e^{-0.577t}$

$e^{-0.0577t}=0.421$

Take natural log on both the sides

- 0.0577 t = - 0.865

t = 15 hours

4 0

3 years ago