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arlik [135]
2 years ago
11

What is the electric field 3.9 m from the center of the terminal of a Van de Graaff with a 6.60 mC charge, noting that the field

is equivalent to that of a point charge at the center of the terminal
Physics
1 answer:
earnstyle [38]2 years ago
3 0

Answer:

the electric field is  3.91 x 10⁶ N/C

Explanation:

Given the data in the question;

Electric field at a point due to point charge is;

E = kq/r²

where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator

Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C

so we substitute into the formula

E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²

E = 59400000 / 15.21

E = 3.91 x 10⁶ N/C

Therefore, the electric field is  3.91 x 10⁶ N/C

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A 12-volt automotive circuit has a current of 3 amps. Technician A says the electric power in this circuit is 36 watts. Technici
skelet666 [1.2K]

Answer:

Technician A is right.                                                    

Explanation:

Given that,

Voltage of circuit, V = 12 volt

Current in the circuit, I = 3 A

Technician A says the electric power in this circuit is 36 watts. Technician B says the electric power in this circuit is 4 watts. We need to say that which technician is correct.

The power of any circuit is given by :

P=V\times I

P=12\ V\times 3\ A

P = 36 watts

So, technician A is right. Hence, this is the required solution.

6 0
2 years ago
Which statement best explains how a planet affects the orbit of a comet as the comet passes by the planet?
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8 0
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daser333 [38]

Answer:

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4 0
3 years ago
A player holds two baseballs a height h above the ground. He throws one ball vertically upward at speed v0 and the other vertica
Degger [83]

Answer:

a)  v = √(v₀² + 2g h),    b)      Δt = 2 v₀ / g

Explanation:

For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.

The velocity of each ball is

ball 1. thrown upwards vo is positive

        v² = v₀² - 2 g (y-y₀)

in this case the height y is zero and the height i = h

        v = √(v₀² + 2g h)

ball 2 thrown down, in this case vo is negative

         v = √(v₀² + 2g h)

The times to get to the ground

ball 1

         v = v₀ - g t₁

         t₁ = \frac{v_{o}  - v }{ g}

ball 2

         v =  -v₀ - g t₂

         t₂ = -  \frac{v_{o}  + v }{ g}  

From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is

       Δt = t₂ -t₁

       Δt = \frac{1}{g} \ [(v_{o} - v)  - ( - v_{o}  - v) ]

       Δt = 2 v₀ / g

6 0
3 years ago
How many Oxygen (O) atoms are in the following? H2O + CO2<br><br> 4<br><br> 3<br><br> 2
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6 0
3 years ago
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