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IRISSAK [1]
3 years ago
7

Suppose you take a short piece of wire that is not attached to anything and move it up and down in a magnetic field. Explain whe

ther or not a current will be induced, and why
Physics
2 answers:
marissa [1.9K]3 years ago
5 0

Answer:

A current will not be induced because the wire is not part of a closed circuit. The electrons have no place to flow to.

Explanation:

Otrada [13]3 years ago
4 0

Answer:

No.

Explanation:

  • According to Faraday's law, the induced emf in the circuit is given by :

         e=\dfrac{d\phi}{dt}, it is proportional to the rate of change of magnetic flux.

  • In this case, a short piece of wire that is not attached to anything and move it up and down in a magnetic field. It means that the circuit is not completed here. It is an open circuit. For the induction of current, a circuit must be completed.
  • Hence, no current will induce.
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What is an example of a wave that is not mechanical and how is it different?
vovikov84 [41]

Answer:

light is an example of a wave that is not mechanical .

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5 0
3 years ago
What is the difference between each distance traveled and displacement travled
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Displacement is a vector magnitude that depends on the position of the body which is individualistic for the trajectory.

While, Distance is a scalar magnitude that measures over the trajectory.

8 0
3 years ago
List 5 reason why water is not thermometric liquid​
kolezko [41]

Answer:

(1) it is transparent so it makes the reading difficult. (2) it is volatile. (3) it is a poor conductor of heat. (4) it has a higher specific heat capacity, so it absorbs more heat from the body with which it is kept in contact. (5) Water cannot be used in thermometer because of its higher freezing point and lower boiling point than other liquids . If water is used in a thermometer , it will start phase change at 0oC and 100oC and will not measure temperature , out of this range .

PLEASE READ CAREFULLY AND PLEASE MARK AS BRAINLIEST :)

5 0
3 years ago
During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±
Alla [95]

Answer:

P = 1 (14,045 ± 0.03 )  k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

            Δ (Pₓ / Py) =?

 Let's start by finding the momentum of each vehicle

car X

        Pₓ = m vₓ

        Pₓ = 2.34 2.5

        Pₓ = 5.85 kg m

car Y

        Py = 2,561 3.2

        Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

          ΔPₓ = m Δv + v Δm

          ΔPₓ = 2.34 0.01 + 2.561 0.01

          ΔPₓ = 0.05 kg m

         ΔP_{y} = m Δv + v Δm

         ΔP_{y} = 2,561 0.01+ 3.2 0.001

         ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

          P = Pₓ / P_{y}

          ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

          ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

          ΔP = 0.006 + 0.0026

          ΔP = 0.009 kg m

The result is

           P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s

7 0
3 years ago
You are designing a delivery ramp for crates containing exercise equipment. The 1470-N crates will move at 1.8 m/s at the top of
alina1380 [7]

Answer:

The force constant ,I = 2394N/m

Explanation:

Given:

Weight of crate,Wg = 1470N

Theta = 22.0°

Kinetic friction,Fk= Fs(max) = 550N

Total length of ramp =8.0m

If y =0 at the bottom of the ramp

y1 = d Sin theta

y1 = 8 × Sin 22°

y1 = 3.0m

y2 = 0

V1=1.8m/s

V2 = 0

The equation combining the gravitational and elastic potential energy, and the work energy theorem is given by:

K1 + Ugrav1 + Uel1 + W = K2 + Uel1 ..eq1

Where KE is given by: 1/2mv^2

Gravitational potential energy is given by: Ugrav = mgy ...eq2

Elastic potential energy = Uel = 1/2Kx^2 ..eq3

The restoring force constant of a spring compressed by a distance x is given by:

Fx = Kx ..eq4

Workdone by a constant force is given by W = Fscostheta ...eq5

Where s = displacement

Theta = the angle between the force and the displacement

Work,W = Wf = 550 ×8 × cos 180°

W = -4400J

From the weight of the crate.mass is:

Wg/g = m

1470/9.8 = 150kg

K1 = 1/2 ×150×1.8^2 = 243m/s

The crate comes to rest at K2=0

Ugrav1 = 150 × 9.8 × 3 = 4410J

Ugrav2 = 150 × 9.8 x 0 = 0J

Uel1= 0 Spring at equilibrium

Substituting the values of the energies and work

253 + 4410 + 0 - 4400 = 0 + 0 + Uel1

Uel1 = 253J

Substituting into eq3

253 = 1/2 Kx^2 = 1/2 Kx(x)

Kx = 506/x

Since crate remains at rest,we use Newton's 2nd law

Fx = fs + Wsin theta

Fx = 550 + 1470 sin 22

Fx = 1100.7N

Substituting into eq4

Kx = 1100.7

X = 506/1100.7 = 0.46m

Kx = 1100.7

K = 1100.7/0.47

K = 2394N/m

5 0
3 years ago
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