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3 years ago

In a motor, electrical current enters through the brushes. a. True b. False

1 answer:

8 0

True, in a motor, electrical current enters through the brushes.
Brush conducts the current from the stationary wires to the moving parts of the motor.
Some examples of this are:
Generator.

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The two vectors and in fig. 3-28 have equal magnitudes of 10.0 m and the angles are 30° and 105°. find the (a) x and (b) y compo

mylen [45]

You can just use basic trigonometry to solve for the x & y components.

vector a = 10cos(30) i + 10sin(30) j = <5sqrt(3), 5>

vector b is only slightly harder because the angle is relative to vector a, and not the positive x-axis. Anyway, this just makes vector b with an angle of 135deg to the positive x-axis.

vector b = 10cos(135) i + 10sin(135) j = <-5sqrt(2), 5sqrt(2)>

So now we can do the questions:

r = a + b

r = <5sqrt(3)-5sqrt(2), 5+5sqrt(2)>

(a) 5sqrt(3)-5sqrt(2)

(b) 5+5sqrt(2)

(c)

|r| = sqrt( (5sqrt(3)-5sqrt(2))2 + (5+5sqrt(2))2 )

= 12.175

(d)

θ = tan-1 ( (5+5sqrt(2)) / (5sqrt(3)-5sqrt(2)) )

θ = 82.5deg

6 0

3 years ago

What is the name of the process where rocks are being broken down into small grains of sediment or soil?

elena-s [515]

Answer:

a. weathering.

Explanation:

weathering is the process of breaking down rocks and minerals into smaller pieces by water, wind, and ice.

4 0

3 years ago

More questions in the image.

Tamiku [17]

True, they represent the direction of motion

8 0

3 years ago

How high can a body vertically thrown with a speed of 40m/s raise after 3 sec (neglecting air

Tcecarenko [31]

y = 75.9 m

Explanation:

y = -(1/2)gt^2 + v0yt + y0

If we put the origin of our coordinate system at the point where a body is launched, then y0 = 0.

y = -(1/2)(9.8 m/s^2)(3 s)^2 + (40 m/s)(3 s)

= -44.1 m + 120 m

= 75.9

5 0

2 years ago

A single loop of wire with an area of 0.0780 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic

Katarina [22]

Answer:

The induced current is 26.7 mA

Explanation:

Given;

area of the loop, A = 0.078 m²

initial magnetic field, B₁ = 3.8 T

change in the magnetic field strength, dB/dt = 0.24 T/s

The induced emf is calculated as;

$emf = - \frac{d \phi}{dt} \\\\emf = -\frac{dB.A}{dt} \\\\emf = A (\frac{dB}{dt} )\\\\emf = 0.078(0.24)\\\\emf = 0.0187 \ V$

The resistance of the loop = 0.7 Ω

The induced current is calculated as;

$V = IR\\\\I = \frac{V}{R} = \frac{emf}{R} = \frac{0.0187}{0.7} = 0.0267 \ A = 26.7 \ mA$

4 0

3 years ago