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ludmilkaskok [199]
3 years ago
7

An electron is brought from rest infinitely far away to rest at point P located at a distance of 0.041 m from a fixed charge q.

That process required 101 eV of energy from an eternal agent to perform the necessary work.(a) What is the electric potential, in volts, at point P?
Physics
1 answer:
torisob [31]3 years ago
3 0

Answer:

The electric potential in volts is 1.618 x 10⁻¹⁷ V

Explanation:

The electric potential, in volts, at point P, can be calculated as follow;

Electric potential is the work done in moving a unit positive charge from infinity to a particular point in the electric field.

Thus, the work done in this process in moving the charge to point p is 101eV.

Convert this Volts = 101 × 1.602 x 10⁻¹⁹ V

                              = 1.618 x 10⁻¹⁷ V

Therefore, the electric potential in volts is 1.618 x 10⁻¹⁷ V

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Answer:

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Explanation:

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3 years ago
a waterwheel built in Hamah, Syria, has a radius of 20 m . If the tangential velocity at the wheels edge is 7.85 m/s , what is t
gtnhenbr [62]

Answer:

3.08m/s²

Explanation:

Given parameters:

Radius = 20m

Tangential velocity  = 7.85m/s

Unknown:

Centripetal acceleration  = ?

Solution:

Centripetal acceleration is the acceleration of a body along a circular path.

 it is mathematically given as;

       a  = \frac{v^{2} }{r}  

v is the tangential velocity

r is the radius

      a  = \frac{7.85^{2} }{20}   = 3.08m/s²

4 0
3 years ago
What two things do you need to know to describe the velocity of an object?
pickupchik [31]
The distance it traveled and the time that it took to travel that distance
3 0
3 years ago
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An object has a force of 88N to the right Wth an acceleration of 8 m/s2 to the right. What is the mass of the object? (mass has
weeeeeb [17]

Answer:

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Explanation:

7 0
3 years ago
Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at se
adoni [48]

Answer:

(a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

(b)\ F_{No} = 771.125N

Explanation:

Given

d_D = 6000ft ---- Altitude of container in Denver

A = 0.0155m^2 -- Surface Area of the container lid

P_D = 79000Pa --- Air pressure in Denver

P_{No} = 100250Pa --- Air pressure in New Orleans

<em>See comment for complete question</em>

Solving (a): The expression for F_{No

Force is calculated as:

F = \triangle P * A

The force in New Orleans is:

F_{No} = \triangle P * A

Since the inside pressure is half the pressure at sea level, then:

\triangle P = P_{No} - \frac{P_{area}}{2}

Where

P_{area} = 101000Pa --- Standard Pressure

Recall that:

F_{No} = \triangle P * A

This gives:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Solving (b): The value of F_{No

In (a), we have:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Where

A = 0.0155m^2

P_{No} = 100250Pa

P_{area} = 101000Pa

So, we have:

F_{No} = [100250 - \frac{101000}{2}] * 0.0155

F_{No} = [100250 - 50500] * 0.0155

F_{No} = 49750* 0.0155

F_{No} = 771.125N

4 0
3 years ago
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