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3 years ago

The magnitude of the electric current is directly proportional to the _____ of the electric field.

2 answers:

8 0

Answer: The magnitude of the electric current is directly proportional to the potential difference of the electric field.

Explanation:

According to Ohm's law, the current flowing between two points in the circuit is directly proportional to the potential difference at a constant temperature.

The electric potential is the amount of work done in moving a unit positive charge from reference point to a particular point in a electric field. And the electric field is equal to the negative potential gradient.

Therefore, the magnitude of the electric current is directly proportional to the potential difference of the electric field.

Sedbober [7]3 years ago

3 0

Is proportional to the CHANGE.

hope this helps

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Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of 2.0A. The dis

Veronika [31]

Answer:

Explanation:

Two straight wires

Have current in opposite direction

i1=i2=i=2Amps

Distance between two wires

r=5mm=0.005m

Length of one wire is ∞

Length of second wire is 0.3m

Force between the wire,

The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by

F/l = μoi1i2/2πr

F/l=μoi²/2πr

μo=4π×10^-7 H/m

The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

F/l = μoi1i2/2πr

F/0.3=4π×10^-7×2²/2π•0.005

F/0.3=1.6×10^-4

Cross multiply

F=1.6×10^-4×0.3

F=4.8×10^-5N

3 0

3 years ago

A bowling ball with a mass of 8 kg is moving at a speed of 5 m/s. What is its<br> kinetic energy?

natali 33 [55]

Answer:

100J

Explanation:

Kinetic energy=1/2mv^2

Kinetic energy=(1/2 x 8)x5^2

Kinetic energy=4x25

Kinetic energy=100

100J

5 0

3 years ago

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$

Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

$\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2} = 4J (6)$

⇒ $\frac{1}{2}* 200N/m* \Delta x_{1} ^{2} = 4J - 3.6 J = 0.4 J (7)$

⇒ $\Delta x_{1} = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)$

6 0

3 years ago

The tongue weight of a trailer should be what percent of the gross trailer weight rating

mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

- Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

7 0

3 years ago

A slab of glass has a 0.600 cm thick layer of water on top of it. A light ray strikes the water at an incident angle of 59.0°. A

Dimas [21]

Answer:

49.63 degree

Explanation:

thickness of glass slab, t = 0.6 cm

angle of incidence = 59 degree

Let r be the angle of refraction

The refractive index of glass, ng = 3/2

refractive index of water, nw = 4/3

refarctive index of glass with respect to water = ng / nw = 3 /2 ÷ 4 /3 = 9 / 8

So, by use of Snell's law

Refractive index of glass with respect to water = Sin i / Sin r

9 / 8 = Sin 59 / Sin r

9 / 8 = 0.857 / Sin r

Sin r = 0.7619

r = 49.63 degree

4 0

3 years ago