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3 years ago

At what location in a circuit is the electrical potential energy the greatest

1 answer:

4 0

It's not the potential energy. It's just the potential.

It's greatest at the positive terminal of the battery or power supply.

You might be interested in

(a) How much work is required to lift a 35-kg object from the ground 3.0 m into the air? (b) How much gravitational potential en

V125BC [204]

Answer:

(a) work required to lift the object is 1029 J

(b) the gravitational potential energy gained by this object is 1029 J

Explanation:

Given;

mass of the object, m = 35 kg

height through which the object was lifted, h = 3 m

(a) work required to lift the object

W = F x d

W = (mg) x h

W = 35 x 9.8 x 3

W = 1029 J

(b) the gravitational potential energy gained by this object is calculated as;

ΔP.E = Pf - Pi

where;

Pi is the initial gravitational potential energy, at initial height (hi = 0)

ΔP.E = (35 x 9.8 x 3) - (35 x 9.8 x 0)

ΔP.E = 1029 J

7 0

3 years ago

The orbit of a certain a satellite has a semimajor axis of 4.0 x 107 m and an eccentricity of 0.15. Its perigee (minimum distanc

Mashutka [201]

Answer:

100KM and 1kkm

Explanation:

5 0

3 years ago

Why do some devices use batteries and other devices use a wall outlet for power?

Alina [70]

Some devices are used while moving around from place to place ... like in the car, on a bicycle, running, or walking the dog. If not for batteries, those devices would need very long extension cords.

3 0

4 years ago

3. A ball is thrown straight up with an initial velocity of 40 m/s.

grandymaker [24]

Answer:

The velocity at the top of its path will be zero (0)

Explanation:

We can solve this problem or particular situation using the principle of energy conservation.

Which tells us that energy is transformed from kinetic energy to potential energy and vice versa. A reference point should be considered at which the potential energy is zero, and at this point the initial velocity of 40 [m/s] is printed to the ball.

$Ek=Ep\\where:\\Ek=kinetic energy [J]\\Ep=potencial energy [J]$

The potential energy is determined by:

$Ep=m*g*h\\where:\\m=mass of the ball[kg}\\g=gravity[m/s^2]\\h=heigth [m]\\$

The kinetic energy is determined by:

$Ek=\frac{1}{2}*m*v_{0} ^{2} \\where\\v_{0} = initial velocity[m/s]$

$Ek=Ep\\\frac{1}{2} *m*v_{0} ^{2} =m*9.81*h\\h=\frac{40^{2}}{2*9.81} \\h=81.5[m]$

This will be the maximum path but, its velocity at this point will be zero. Because now all the kinetic energy has been transformed in potential energy.

8 0

4 years ago

Outside the space shuttle, you and a friend pull on two ropes to dock a satellite whose mass is 700 kg. The satellite is initial

MaRussiya [10]

Answer:

your friend did a work of 1717.34 J

Explanation:

The sum of the work made by your friend and you is equal the change in the kinetic energy, so:

$W_a+W_b=\frac{1}{2}mv_f^{2}-\frac{1}{2}mv_i^{2}$

Where $W_a$ is the work that you did, $W_b$ is the work that your friend did, m is the mass of the satellite and $v_f$ and $v_i$ are the final and initial speeds