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ololo11 [35]
3 years ago
14

A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S

uppose the magnetic field B begins to decrease rapidly in strength
Requried:
What happens to the loop?

1. The loop is pushed to the left, toward the magnetic field.
2. The loop doesn’t move.
3. The loop is pushed downward, towards the bottom of the page.
4. The loop will rotate.
5. The loop is pushed upward, towards the top of the page.
6. The loop is pushed to the right, away from the magnetic field
Physics
1 answer:
Black_prince [1.1K]3 years ago
8 0

Answer:

. The loop is pushed to the right, away from the magnetic field

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

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A roller-coaster car has a mass of 1040 kg when fully loaded with passengers. As the car passes over the top of a circular hill
rusak2 [61]

Answer:

a.6373.5 N

b.-3837.6 N

Explanation:

Mass of roller coaster=m=1040 kg

Radius=r=24 ,

a.v=9.4m/s

Normal force=F_N

According to question

mg-F_N=\frac{mv^2}{r}

Where g=9.81 m/s^2

Substitute the values

1040\times 9.81-F_N=\frac{1040\times (9.4)^2}{24}

10202.4-F_N=3828.9

F_N=10202.4-3828.9=6373.5 N

b.v=18m/s

g=9.81 m/s^2

1040\times 9.81-F_N=\frac{1040\times (18)^2}{24}

10202.4-F_N=14040

F_N=10202.4-14040=-3837.6 N

8 0
4 years ago
A 52 kg child on a swing is travelling at 6 m/s . What is his gravitational potential energy if he has 1000 J of the mechanical
DiKsa [7]

Answer:

The correct answer is "64 J".

Explanation:

The given values are:

Mass,

m = 52 kg

Velocity,

v = 6 m/s

Mechanical energy,

= 1000 J

Now,

The gravitational potential energy will be:

⇒ P.E=1000-\frac{1}{2}mv^2

           =1000-\frac{1}{2}\times 52\times (6)^2

           =1000-26\times 36

           =1000-936

           =64 \ J

7 0
3 years ago
A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi
Darina [25.2K]

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

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Let this oscillator have the same energy as a mass on a spring, with the same k and m, released from rest at a displacement of 5
allochka39001 [22]
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I think it would need to decress bro

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