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2 years ago

What is electromagnetic induction?

Physics

2 answers:

dlinn [17]2 years ago

4 0

Answer:

Electromagnetic or magnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.

Explanation:

correct me if im wrong:)

gulaghasi [49]2 years ago

3 0

The phenomenon of generation of current or emf by changing the magnetic flux is known as Electromagnetic Induction.

First Law Whenever magnetic flux linked with the closed loop or circuit changes, an emf induces in the loop or circuit which lasts so long as change in flux continuous.

Explanation:

Hope it helps~

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A cube with 30-cm-long sides is sitting on the bottom of an aquarium in which the water is one meter deep. (Round your answers t

OleMash [197]

Answer:

A) hydrostatic force on top of cube = 882.9N

B) hydrostatic force on sides of cube = 0N

Explanation:

Detailed explanation and calculation is shown in the image below

4 0

3 years ago

Water flows from the bottom of a storage tank at a rate of r(t) = 300 − 6t liters per minute, where 0 ≤ t ≤ 50. Find the amount

SSSSS [86.1K]

r(t) models the water flow rate, so the total amount of water that has flowed out of the tank can be calculated by integrating r(t) with respect to time t on the interval t = [0, 35]min

∫r(t)dt, t = [0, 35]

= ∫(300-6t)dt, t = [0, 35]

= 300t-3t², t = [0, 35]

= 300(35) - 3(35)² - 300(0) + 3(0)²

= 6825 liters

7 0

3 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago

Two point charges 3q and −8q (with q > 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so

riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q 8q / L²

Q = 8q (x² / L²)

so charge required = - 8q (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

3 0

3 years ago

An 1800-kg truck pulls a 710-kg trailer away from a stoplight with an acceleration of 1.20 m/s2 . Part A What is the net force e

faust18 [17]

Answer:

F = 852 N

Explanation:

We apply Newton's second law to the trailer :

F = m*a Formula (1)

F : net force exerted by the truck on the trailer Newtons (N)

m : mass of the trailer in kilograms (kg)

a : acceleration of the trailer in meters over second square (m/s²)

Data

a=1.20 m/s² : acceleration of the trailer

m=710 kg : mass of the trailer

We replace data in the Formula (1) to calculate the net force exerted by the truck on the trailer

F = (710 kg)*(1.20 m/s²)

F = 852 N

8 0

3 years ago

What is electromagnetic induction?​

What is electromagnetic induction?