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Vika [28.1K]
3 years ago
8

A magnetic field points up through a loop where the area can be increased by dragging the side of the loop along the rails that

make up 2 of the other sides. If you are looking down on the loop (magnetic field out of the page from your perspective) what is the direction of the induced current as the loop increases in area
Physics
1 answer:
Aliun [14]3 years ago
5 0

Answer:

Explanation:

The magnetic field is pointing upwards , so magnetic flux is upwards . Due to increase in area , magnetic flux in upward direction will increase . Hence according to Lenz's law , induced current will be such that it will try to counter this increase . It is possible if current is in clockwise direction . Due to it magnetic field will be induced in downward direction thus reducing the total magnetic flux.  

Hence direction of induced current will be clockwise direction in the loop as shown in the image enclosed .

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Which of the following are early clues that signal an earthquake may occur? Select all that apply. Changes in magnetic propertie
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The following answers apply;

  • Changes in magnetic properties of rock
  • Decrease in well water levels
  • Increases in radon gas in groundwater
  • Foreshocks  

These other choices may be good indicators of an imminent volcanic eruption;

Movement of magma

Increase in sulfur dioxide and carbon dioxide ground emissions


6 0
3 years ago
Read 2 more answers
2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he
Ratling [72]

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, v_s = d/t₂

∴ v_s = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

5 0
3 years ago
g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe
olga_2 [115]

Answer:

The minimum compression is  x= 0.046m

Explanation:

From the question we are told that

              The mass of the block is m_b = 1.45 kg

               The spring constant is  k = 860 N/m

               The coefficient of static friction is  \mu = 0.36

For the the block not slip it mean the sum of forces acting on the  horizontal axis is equal to the forces acting on the vertical axis

     Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

                   F_y = m_b*g

And the force acting on the horizontal axis is  force due to the spring which is mathematically represented as

                   F_x = k *x * \mu

where x is the minimum compression to keep the block from slipping

        Now equating this two formulas and making x the subject

                      x = \frac{m_b * g}{k * \mu}

substituting values we have

                     x = \frac{1.45 * 9.8}{860 *0.36}

                        x= 0.046m

 

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3 years ago
A copper block rests 30.0 cm from the center of a steel turntable. The coefficient of static friction between the block and the
PIT_PIT [208]

Answer:

refer to the above attachment

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1 year ago
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Answer:

Answer: 2.70m/s NE

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