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sesenic [268]
3 years ago
14

Hi..I'm new here..Can some one help mi with my school work plz​

Physics
1 answer:
kirill [66]3 years ago
3 0

Answer:

sure if its easy im jp ill help u

Explanation:

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A source charge generates an electric field of 4286 N/C at a distance of 2. 5 m. What is the magnitude of the source charge? (Us
svp [43]

The magnitude of the source charge is 3 μC which generates 4286 N/C of the electric field. Option B is correct.

What does Gauss Law state?

It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.

Q = \dfrac {ER^2}k

Where,

E = electric force = 4286 N/C

k = Coulomb constant = 8.99 \times  10^9 \rm\ N m ^2 /C ^2

Q\\
     = charges = ?

r = distance of separation = 2.5 m

Put the values in the formula,

Q  = \dfrac {4286\times  2.5 ^2}{8.99 \times  10^9 }\\\\
Q  = 3\rm \  \mu C

Therefore, the magnitude of the source charge is 3 μC.

Learn more about Gauss's law:

brainly.com/question/1249602

8 0
2 years ago
Black holes must have a mass three or more times the mass of _____.
tia_tia [17]
The answer is <span>d. the sun</span>
7 0
3 years ago
Read 2 more answers
Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th
Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of aluminium = 0.90J/g^oC

c_2 = specific heat of water = 4.184J/g^oC

m_1 = mass of aluminum = 0.500 kg = 500 g

m_2 = mass of water = 0.250 kg  = 250 g

T_f = final temperature of mixture = ?

T_1 = initial temperature of aluminum = 150^oC

T_2 = initial temperature of water = 20^oC

Now put all the given values in the above formula, we get:

500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC

T_f=59.10^oC

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 59.10^oC

8 0
3 years ago
An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. The acceleration
Debora [2.8K]

Answer:

The radius of the loop is  20.66 km

Explanation:

let the radius of the loop be r

mass of airplane is m

At the top, the pilot experiences two radial forces, which are

1) Gravitational force is  mg

2) Centrifugal forces mv²/r out of the center

When the pilot experiences no weight,

then, mg = mv²/r

r = v² / g

 = 450² / 9.8

 = 20.66 x 10³3

 = 20.66 km

3 0
3 years ago
Consider two wind turbines, each located in sites with the same wind speed, and each with the same efficiency. Turbine 1 has a r
Alex787 [66]

Answer:

shut up nerd

Exption:

7 0
3 years ago
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