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Elenna [48]
3 years ago
9

If a charge is located at the center of a spherical volume and the electric flux through the surface of the sphere is φ, what wo

uld be the flux through the surface if the radius of the sphere were tripled?
Physics
1 answer:
Svetradugi [14.3K]3 years ago
3 0

the correct answer is if you tripled the sphere volume, do the mat idiot

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Three thermometers are in the same water bath. After thermal equilibrium is established, it is found that the Celsius thermomete
Nikolay [14]

Answer:

OPTION A, Kelvin Thermometer is Incorrect

Explanation:

Now, if you consider best two out of three results, then celsius and Fahrenheit thermometers read the same value, meaning both are right.

1) K = °C + 273

K = 100°C + 273

k = 373°C

Kelvin Thermometer is Incorrect

2) C = \frac{F - 32}{1.8}

when we have 212°F

C = \frac{212 - 32}{1.8} \\\\= 100^\circ C

which is correct

3 0
3 years ago
If a cart of 10 kg mass has a force of 5 newtons exerted on it, what is its acceleration? m/s2
Elza [17]

Answer:

= 0.5 m/s²

Explanation:

  • According to Newton's second law of motion, the resultant force is directly proportion to the rate of change of linear momentum.

Therefore;<em> F = ma , where F is the Force, m is the mass and a is the acceleration.</em>

<em>Thus; a = F/m</em>

<em>but; F = 5 N, and m = 10 kg</em>

<em>  a = 5 /10</em>

    <u>= 0.5 m/s²</u>

4 0
3 years ago
Read 2 more answers
Work done depends on
natima [27]

Answer:

C. Both force and displacement

Explanation:

Hope this helps

3 0
3 years ago
The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A
vazorg [7]

Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

8 0
3 years ago
A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p
alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

F=qvB

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

qvB = \frac{mv^2}{r}

which can be rewritten as

v=\frac{qB}{mr}

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

\frac{2\pi r}{T}=\frac{qB}{mr}

So, we get:

T=\frac{2\pi m r^2}{qB}

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) a. f/10

The frequency of revolution of a particle in uniform circular motion is

f=\frac{1}{T}

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}

6 0
3 years ago
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