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Pavel [41]
3 years ago
13

Suppose you were asked to demonstrate electromagnetic induction. Which of the following situations will result in an electric cu

rrent?
A. A magnet is moving toward a wire loop.
B. A wire loop is moving away from a magnet.
C. A wire loop is rotated near a magnet.
D. All of the above
Physics
2 answers:
steposvetlana [31]3 years ago
5 0
D. All of the above. When a wire loop is moved or rotated in a magnetic field, there is a change in magnetic flux which produces emf in wire loop and hence electric current is produced. 
ad-work [718]3 years ago
3 0
The answer is D. It is obvious since it is a one-choice question and both A and B state the exact same thing. Therefore the answer has to be D and so C has to be correct as well. Exam techniques! :D
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How would the Earth move if the sun (including its gravity) suddenly disappeared? Explain your answer.
daser333 [38]
It would float outward as all space is expanding from the center constantly. of by some chance it for caught in the orbit of another celestial body with huge mass it may orbit that, bur other wise it would just push outward from the center of the universe theoretically the big bang
5 0
3 years ago
Displacement vectors of 10 m west and 14 m west make a resultant vector that is ?
goldenfox [79]

Resultant vector= 24 m west.

Explanation:

resultant vector= r1+ r2 when the two vectors are in same direction

r1=10m west

r2= 14 m west

so resultant displacement= 10 + 14

resultant displacement= 24 m west

4 0
2 years ago
The impulse given to a ball with mass of 4 kg is 28 N s. If the ball were already moving at 3 m/s what would the final velocity
mart [117]

vf = 10 m/s. A ball with mass of 4kg and a impulse given of 28N.s with a  intial velocity of 3m/s would have a final velocity of 10 m/s.

The key to solve this problem is using the equation I = F.Δt = m.Δv, Δv = vf - vi.

The impulse given to the ball with mass 4Kg is 28 N.s. If the ball were already moving at 3 m/s, to calculate its final velocity:

I = m(vf - vi) -------> I = m.vf - m.vi ------> vf = (I + m.vi)/m ------> vf = I/m + vi

Where I 28 N.s, m = 4 Kg, and vi = 3 m/s

vf = (28N.s/4kg) + 3m/s = 7m/s + 3m/s

vf = 10 m/s.

.

6 0
3 years ago
A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t = 0 the spring is neither stretch
r-ruslan [8.4K]

Answer:

a) A = 0.98 m

b) Ф = 90°

c) x = -0.98sin(12.25t)

Explanation:

We know the value of the spring constant which is 300 N/m, the innitial apmplitude, that we will call it xo is 0, at t = 0, and the speed is 12 m/s

The expression for the amplitude under these conditions is:

A = √xo² + vx²/w² (1)

To calculate the angular speed w, we use the following expression:

w = √k/m  (2)

Calculating w:

w = √300/2 = 12.25 rad/s

Now, we replace this value into equation 1, along with the other known values and solve for A:

A = √0 + (12)²/(12.25)²

A = 0.98 m

b) In this part, is actually easy, the displacement of x in function of the time is given by:

x = A cos(wt - Ф) (4)

But at t = 0 we have then:

x = xo = A cosФ (5)

Solving for the angle Ф we have:

xo/A = cosФ

Ф = arccos(x0/A)  (6)

Replacing the data in (6):

Ф = arccos(0/0.98)

Ф = 90°

c) Equation (4) is the expression for the simple harmonic motion

x = A cos(wt - Ф)

And if we replace here the value of w and the previous angle, we can write an equation for x in function of t:

x = 0.98 cos (12.25t - 90)

x = 0.98 cos(12.25t - 90)

And we have an trigonometric expression for cos that is:

cos(α - π/2) = -sinα

in this case, α will be the value of w = 12.25 rad/s. and 90° is the same as rewritting as π/2, therefore:

cos(α - π/2) = cos(12.25t - 90)

x = -0.98sin(12.25t)

3 0
2 years ago
A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

8 0
2 years ago
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