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Ksivusya [100]
4 years ago
6

A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

the electric potential energy of the point charge?
________ × 10–4 J
What is the electric potential energy of the point charge at 1.3 m?

_________ × 10–3 J
Physics
2 answers:
schepotkina [342]4 years ago
3 0
4 for the first one and 5.2 for the second one.
natali 33 [55]4 years ago
3 0

E = electric field = 1 x 10³ N/C

q = charge = 4 x 10⁻⁶ C

d = distance = 0.10 m

electric potential energy is given as

U = q E d

inserting the above values

U = (4 x 10⁻⁶ ) (1 x 10³) (0.1)

U = 4 x 10⁻⁴ J


when distance = d = 1.3 m

electric potential energy is given as

U = q E d

inserting the above values

U = (4 x 10⁻⁶ ) (1 x 10³) (1.3)

U = 5.2 x 10⁻³ J

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Determine the shortest frequency of light required to remove an electron from a sample of ti metal if the binding energy of tita
Mama L [17]
The correct answer is: shortest frequency = 7.86*10^15 s^{-1}

Explanation: 

The binding energy of titanium = 3.14*10^6<span> J/mol
</span>The energy required to remove an electron = (3.14*10^6) /(6.023*10^23) = 5.213*10^-18 J<span>
Where 6.023*10^</span>23 = Avagadro number

Since E = hv

Frequency = v = E/h
E = Energy = 5.213*10^-18
h = Planck's constant = 6.626*10^-34
v = (5.213*10^-18) / 6.626*10^-34<span>)
</span><span>v = </span><span>7.86*10^15 </span>s^{-1} (shortest frequency)
7 0
4 years ago
I need help please...
Natalka [10]
No cluuuueee :/ sowwwwyyy but good luck
6 0
3 years ago
Read 2 more answers
Please help me on my physics hw ​
MatroZZZ [7]

Answer:

1. 11 A

2. 240 V

3. 8 Ω

4. 60 C

5. 14400 C

Explanation:

1. Determination of the current.

Voltage (V) = 110 V

Resistance (R) = 10 Ω

Current (I) =?

V = IR

110 = I × 10

Divide both side by 10

I = 110 / 10

I = 11 A

2. Determination of the voltage

Current (I) = 3 A

Resistance (R) = 80 Ω

Voltage (V) =?

V = IR

V = 3 × 80

V = 240 V

3. Determination of the resistance.

Current (I) = 0.5 A

Voltage (V) = 4 V

Resistance (R) =?

V = IR

4 = 0.5 × R

Divide both side by 0.5

R = 4 / 0.5

R = 8 Ω

4. Determination of the charge

Current (I) = 2 A

Time (t) = 30 s

Charge (Q) =?

Q = it

Q = 2 × 30

Q = 60 C

5. Determination of the charge.

We'll begin by converting 20 mins to seconds. This can be obtained as follow:

1 min = 60 s

Therefore,

20 mins = 20 × 60

20 mins = 1200 s

Finally, we shall determine the charge as follow:

Current (I) = 12 A

Time (t) = 1200 s

Charge (Q) =?

Q = it

Q = 12 × 1200

Q = 14400 C

6 0
3 years ago
A boy pushes a 54kg cart with a force of 81N. What is the cart's acceleration?
borishaifa [10]

Explanation:

maturity also get disorganised

8 0
3 years ago
Earth has a magnetic dipole moment of 8.0 × 1022 J/T. (a) What current would have to be produced in a single turn of wire extend
Bogdan [553]

Answer:

a

The current that would be produced is I = 6.26 *10 ^8 A

b

Yes this arrangement can be used to cancel out earths magnetic field at points well above the Earth's surface.This is because this current loop acts as a magnetic dipole for point above the earth surface  

c

No this arrangement can not be used to cancel out earths magnetic field at points on the Earth's surface .this because on the earth surface it shifts from its behavior as a magnetic dipole

Explanation:

    From the question we are told that

             The magnetic moment of earth is M = 8.0*10 ^{22} J/T

               The radius of earth generally has a value of R = 6378 *10^3 m

Magnetic moment is mathematically given as

                    M = IA

A is the area of the of the earth(assumption that the earth is circular ) and this evaluated as  

                     A = \pi R^2

Now making I the subject in the above formula

                  I = \frac{M}{A}

                     = \frac{M}{\pi R^2}

                     = \frac{8.0^10^{22}}{\pi (6378 *10^{3})^2}

                     = 6.26 *10^8 A

                   

5 0
3 years ago
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