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Genrish500 [490]
2 years ago
13

Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele

ctric field of N/C at point P. Determine the work required of an external agent to move a 3.0-μC charge along the x axis from infinity to point P without any net change in the kinetic energy of the particle.
Physics
1 answer:
Julli [10]2 years ago
4 0

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C is the magnitude of the charge

\Delta V = 6.0 kV = 6000 V is the potential difference between point P and infinity

Substituting into the equation, we find

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

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Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th
katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

3 0
2 years ago
Read 2 more answers
A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t
Reil [10]

Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0
2 years ago
A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an
Cloud [144]

Answer:

1.36

Explanation:

n_{air} = Index of refraction of air = 1

n_{plastic} = Index of refraction of plastic = ?

i = angle of incidence in air = 32.0° deg

r = angle of refraction in plastic = 23.0° deg

Using Snell's law

n_{air} Sini = n_{plastic} Sinr

(1) SIn32.0 = n_{plastic} Sin23.0

n_{plastic} = 1.36

5 0
2 years ago
An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele
leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

\alpha = \frac{\omega_f - \omega_0}{t}

Here

\alpha = Angular acceleration

\omega_{f,i} = Initial and final angular velocity

There is not initial angular velocity,then

\alpha = \frac{\omega_f}{t}

We know that the relation between the tangential velocity with the angular velocity is given by,

v = r\omega

Here,

r = Radius

\omega = Angular velocity,

Rearranging to find the angular velocity

\omega = \frac{v}{r}}

\omega = \frac{30}{0.20} \rightarrow Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

\alpha = \frac{30}{0.20*6}

\alpha = 25 rad /s^2

Therefore teh average angular acceleration of each wheel is 25rad/s^2

3 0
2 years ago
A spring with a spring constant of 0.70 N/m is stretched 1.5 m. What was the force?
Talja [164]

Answer:

1.05 N

Explanation:

K = 0.7 N/m

e = 1.5 m

F = ?

from Hooke's law of elasticity

F = Ke

= 0.7×1.5

= 1.05 N

5 0
2 years ago
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