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Dovator [93]
3 years ago
5

Increasing which of the following would increase the magnetic force between the permanent magnet and the coil? A. Transformer B.

Circuit C. Current D. Generator
Physics
1 answer:
Gnoma [55]3 years ago
7 0

For this one, all you really need to do is eliminate any answers
that are absurd or meaningless.

You can't increase a transformer.
You can't increase a circuit.
You can't increase a generator.

When the <em><u>current</u></em> through a coil of wire increases, the magnetic field
around the coil increases, so there would be more magnetic force
between the coil and a permanent magnet.


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grigory [225]
Rocks, earth aging, fossils
3 0
3 years ago
A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s. If the car was in contact with the wall
Alinara [238K]

Answer:

<h2>57166.6N</h2>

Explanation:

Step one:

given data

mass=700kg

initial velocity u=29m/s

final velocity=4.5m/s

time t= 0.3second.

Step two:

The expression for the momentum is given as

FΔt=mΔv

make F subject of the formula

F=mΔv/Δt

Substitute our given data we have

F=700(29-4.5)/0.3

F=700*24.5/0.3

F=17150/0.3

F=57166.6N

the force is 57166.6N

6 0
3 years ago
47. the beam is supported by two rods ab and cd that have cross-sectional areas of 12mm^2 and 8mm^2, respectively. determine the
Ugo [173]

Let the beam is of length L

Now the stress on both the end is same

now we can say that torque on the beam due to two forces must be zero

N_1* x =  N_2* (L - x)

also we know that stress at both ends are same

\frac{N_1}{12} = \frac{N_2}{8}

2*N_1 = 3*N_2

Now from two equations we have

\frac{3}{2}N_2*x = N_2* (L - x)

solving above equation we have

x = \frac{2}{5}L

<em>so the load is placed at distance 0.4L from the end of 12 mm^2 area</em>

8 0
3 years ago
Which of these would have a volume equal to about 2 cm³?
postnew [5]
Answer could be2 grains of rice
5 0
3 years ago
A car travels along a straight road, heading east for 1 h, then traveling for 30 min on another road that leads northeast. If th
Bogdan [553]

Answer:

The car is 72.75 miles away from its starting position.

Explanation:

First, remember the relation:

distance = time*speed.

Also, the distance between two points (a, b) and (c, d) is:

D = √( (a - c)^2 + (b - d)^2)

For this problem, we can assume:

The North is equivalent to the y-axis, and the East is equivalent to the x-axis.

We also assume that the initial position of the car is (0mi, 0mi)

Now the car moves to the East at a speed of 52mi/h for one hour, then the new position of the car is:

(0mi, 0mi) + (52mi/h*1h, 0mi) = (52mi, 0mi)

Now the car travels 30 mins (or 0.5 hours) to the northeast at a speed of 52mi/h.

We can assume that it moves at an exact angle of 45° from East to North, then the components of the speed can be written as:

Sx = speed in the x-axis = 52mi/h*cos(45°) = 36.77 mi/h

Sy = speed in the y-axis = 52mi/h*sin(45°) =  36.77 mi/h

Then the new position of the car is:

(52mi, 0mi) + (36.77 mi/h*0.5h, 36.77 mi/h*0.5h) = (70.385 mi, 18.385 mi)

Now we know the final position of the car.

The distance between the final position (70.385 mi, 18.385 mi) and the initial position (0mi, 0mi) is:

D = √( (70.385 mi - 0mi)^2 + (18.385 mi - 0mi)^2) = 72.75 mi

The car is 72.75 miles away from its starting position.

7 0
3 years ago
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