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2 years ago

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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A 1400kg car is moving at a speed of 25m/s. How much KE does the car have?

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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]

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v=115 m/s

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

Solution

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

For speed in km/h(kilometers per hour)

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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