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3 years ago

7. How much work is done in moving a charge of 10 micro coulombs 1 meter along an equipotential of 10 volts?

2 answers:

8 0

It takes work to push charge through a change of potential.
There's no change of potential along an equipotential path,
so that path doesn't require any work.

Neporo4naja [7]3 years ago

7 0

0, since moving along an equipotential requires no work.

You might be interested in

The temperature dependence of classical expression for electrical resistivity is

Gemiola [76]

Answer:

The resistivity of materials depends on the temperature as ρt = ρ0 [1 + α (T – T0). This is the equation that shows the relationship between the resistivity and the temperature.

hope it helps

5 0

3 years ago

A 50.0 ohm and a 30.0 ohm resistor are connected in parallel. What is their equivalent resistance? Unit=Ohms

Alchen [17]

R(parallel) = product/ sum

50×30/50+30

1500/80

18,75 ohms

4 0

3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

The radial acceleration is $a_c = 0.9574 m/s^2$

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

A block of mass m = 4.8 kg slides from left to right across a frictionless surface with a speed Vi=7.3m/s. It collides with a bl

nirvana33 [79]

Answer:

The speed of the 11.5kg block after the collision is V≅4.1 m/s

Explanation:

ma= 4.8 kg

va1= 7.3 m/s

va2= - 2.5 m/s

mb= 11.5 kg

vb1= 0 m/s

vb2= ?

vb2= ( ma*va1 - ma*va2) / mb

vb2= 4.09 m/s ≅ 4.1 m/s

7 0

3 years ago

Two identical bullets are used. Both are released at the same height - one fired out of a gun, the other is dropped. Ignoring ai

irina [24]

Answer:

Both bullets will hit the ground at the same time.

Explanation:

Let's only analyze the vertical problem.

Any object that is not in the floor or resting in some site is being affected by the gravitational force (remember that we are ignoring air resistance)

Then the acceleration of this object will be equal to the gravitational acceleration:

a = -9.8m/s^2

Where the minus sign is because this acceleration goes down.

To get the velocity equation we need to integrate over time, we will get:

v(t) = ( -9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity.

To get the position equation we need to integrate over time again, we will get:

p(t) = (1/2)*( -9.8m/s^2)*t^2 + v0*t + H

Where H is the initial height.

p(t) = (-4.9 m/s^2)*t^2 + v0*t + H

The object will hit the ground when p(t) = 0

Then we need to solve for t the next equation:

(-4.9 m/s^2)*t^2 + v0*t + H = 0

Notice that the only things we need to know are:

H = initial height (we know that is the same for both bullets)

v0 = initial vertical velocity (also is the same for both bullets)

Notice that the horizontal velocity does not affect this equation, then we will get the same value of t for the dropped bullet and for the fired bullet.

This means that both bullets will hit the ground at the same time.

8 0

3 years ago