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stepan [7]
2 years ago
15

How much work was done on a charge of 3.0 to move it through 12v electric potential difference

Physics
2 answers:
erik [133]2 years ago
8 0

Answer:

  • Hence, the work done in moving the charge is 45 J.

<h2>( Too lazy to translate it to Spanish T_T )</h2>
Ray Of Light [21]2 years ago
8 0

Answer:

45 J

Explanation:

Potential difference (V)= 15 V

Charge (Q)= 3C

We know that

The potential difference(p.d) between two point in an electric circuit is defined as the amount of work done in moving a unit charge from one point to the other point.

Potential difference (V) = work done (W)/Charge (Q)

V= W/Q

Putting the values in the above formula we get:

15 V = W / 3 C

W= 15 V × 3 C

W= 45 J

Hence, the work done in moving the charge is 45 J.

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Temperatures of gases inside the combustion chamber of a four‑stroke automobile engine can reach up to 1000°C. To remove this en
Alona [7]

Answer:

thats a lot, which one u want me to do?

Explanation:

8 0
3 years ago
What is an instrument commonly used to measure wind speed?
xeze [42]
The correct answer is letter D. Anemometer. It is a device that is used to measure wind speed. It is a very common weather station instrument and is available to use and to make. Anemos, from the greek word that means wind.
5 0
3 years ago
Read 2 more answers
A 1400kg car is moving at a speed of 25m/s. How much KE does the car have?
Nady [450]

Answer:

437500Joules

Explanation:

Kinetic energy=1/2mvsquare

1/2 x 1400 x 25 x25

kinetic energy= 437500Joules

6 0
3 years ago
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

A_{area}=0.140m^{2}\\  p_{air}=1.21 kg/m^{3}\\  m_{mass}=80kg

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

mg=1/2*C*p_{air}*v^{2}*A_{area}\\   v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

v=(115*3.6)km/h\\v=414km/h

5 0
3 years ago
An object of mass m = 5.0 kg hangs from a cord around a light pulley: The length of the cord between the oscillator and the pull
puteri [66]

Answer:

\mu=0.0049Kg/m

Explanation:

When a standing wave is formed with six loops means the normal mode of the wave is n=6, the frequency of the normal mode is given by the expression:

f_n=\frac{nv}{2L}

Where L is the length of the string and v the velocity of propagation. Use this expression to find the value of v.

f_6=\frac{6v}{2L}\\(150)=\frac{6v}{2(2)} \\150=\frac{3v}{2} \\3v=150(2)\\ v=\frac{300}{3} \\v=100m/s

The velocity of propagation is given by the expression:

v=\sqrt{\frac{T}{\mu }

Where \mu is the desirable variable of the problem, the linear mass density, and T is the tension of the cord. The tension is equal to the weight of the mass hanging from the cord:

T=W=mg=(5)(9.81)=49.05N

With the value of the tension and the velocity you can find the mass density:

v=\sqrt{\frac{T}{\mu}

v^2=\frac{T}{\mu}\\ \mu=\frac{T}{v^2} =\frac{49.05}{(100)^2} =\frac{49.05}{10000} =0.0049Kg/m

6 0
3 years ago
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