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Anni [7]
2 years ago
3

Which of the following laws may be used to find the direction of the current induced in a loop close to a moving magnet?

Physics
2 answers:
Irina-Kira [14]2 years ago
8 0
According to Lenz's law, the induced current produces a magnetic filed that opposes the changing magnetic field producing it. For example if the magnet is getting closer to the loop, the current produced in the loop must be such that its magnetic field pushes back the approaching magnet. If the magnet is moving away from the loop, the magnetic field due to the induced current must attract the magnet. 
<span>Answer: D</span>
ira [324]2 years ago
7 0
The answer is d


have fun :))))))
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A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass
Basile [38]

a) 120 s

b) v = 0.052R [m/s]

Explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

L=2\pi R

where R is the radius of the wheel.

The period of revolution is:

T=120 s

Therefore, the tangential speed of the book is:

v=\frac{L}{T}=\frac{2\pi R}{120}=0.052R

8 0
2 years ago
Gravity is the force that keeps us on the Earth. It pulls us towards the center of the Earth. If you were to move from the surfa
Nikitich [7]
<h2>Answer: B. Gravitational potential energy </h2>

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>

That is why this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the <u>Earth</u>, in which  <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy U_{p} will be:

U_{p}=mgh  

Where m is the mass of the object, g the acceleration due gravity and h the height of the object.

As we can see, the value of U_{p} is directly proportional to the height.

6 0
2 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
2 years ago
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfo
dolphi86 [110]

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³  / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

3 0
2 years ago
The late news reports the story of a shooting in the city. Investigators think that they have recovered the weapon and they run
Gemiola [76]

Answer:

50000 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of bullet = 0.050 kg

velocity (v) = 400 m/s

Distance (s) = 0.080 m

Force (F) =?

Next, we shall determine the acceleration of the bullet. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 400 m/s

Distance (s) = 0.080 m

Acceleration (a) =?

v² = u² + 2as

400² = 0 + (2 × a × 0.08)

160000 = 0 + 0.16a

160000 = 0.16a

Divide both side by 0.16

a = 160000 / 0.16

a = 1×10⁶ m/s²

Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:

Mass (m) of bullet = 0.050 kg

Acceleration (a) of bullet = 1×10⁶ m/s²

Force (F) =?

F = ma

F = 0.050 × 1×10⁶

F = 50000 N

Thus, the bullet exerted a force of 50000 N on the target.

7 0
2 years ago
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