All categories

3 years ago

Which of the following would cause electromagnetic induction

1 answer:

5 0

<span>It occurs whenever a magnetic field and an electric conductor move relative to one another so the conductor crosses lines of force in the magnetic field. The current produced by electromagnetic induction is greater when the magnet or coil moves faster, the coil has more turns, or the magnet is stronger</span>

You might be interested in

HELP ME PLEAASSEEE ILL MARK U THE BRAINLIEST

Alika [10]

Answer:

Gravitational energy and kinetic energy

7 0

3 years ago

A metal coin has a volume of 835 mm3 and a mass of 5.67 g. What is the density of the coin?

Lilit [14]

Density formula: d = M/V
So I think the answer would be 5.67/ 835 I am not sure of the answer I got confused (´∀｀) but I hope it will help

3 0

3 years ago

Read 2 more answers

A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th

ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

5 0

3 years ago

Which model is most likely used to predict a future event?

snow_lady [41]

Predictive modeling.

4 0

3 years ago

If the magnitude of the electric field in air exceeds roughly 3 âœ• 106 n/c, the air breaks down and a spark forms. for a two-di

Vlad1618 [11]

Answer: 39.8 μC

Explanation:

The magnitude of the electric field generated by a capacitor is given by:

$E = \frac{V}{d}$

d is the distance between the plates.

For a capacitor, charge Q = CV where C is the capacitance and V is the voltage.

$C =\frac{\epsilon_o A }{d}$

where A is the area of the plate and ε₀ is the absolute permittivity.

substituting, we get

$E = \frac{Q}{\epsilon_o A}$

It is given that the magnitude of the electric field that can exist in the capacitor before air breaks down is, E = 3 × 10⁶ N/C.

radius of the plates of the capacitor, r = 69 cm = 0.69 m

Area of the plates, A = πr² = 1.5 m²

Thus, the maximum charge that can be placed on disks without a spark is:

Q = E×ε₀×A

⇒ Q = 3 × 10⁶ N/C × 8.85 × 10⁻¹² F/m × 1.5 m² = 39.8 × 10⁻⁶ C = 39.8 μC.

8 0

3 years ago