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Svetach [21]
3 years ago
10

A circular coil has a 10.0 cm radius and consists of 30 closely wound turns of wire. An externally produced magnetic field of 2.

6 mT is perpendicular to the plane of the coil.
(a) if there is no current in the loop, what is the magnetic flux through the coil?
(b) When the current in the coil is 3.8 A in certain direction, the net flux through the coil is found to vanish. What is the inductance of the coil
Physics
1 answer:
sergey [27]3 years ago
6 0

Answer:

(a) 0.00245 Wb

(b) 0.000645 H

Explanation:

(a) The magnetic flux, Ф, through a coil of radius, r, having N turns due to a magnetic field, B, is given as:

Ф = NBA

where A = area of coil = \pi r^2

Ф = NB\pi r^2

Ф = 30 * 2.6 * 10^{-3} * \pi  * 0.1^2

Ф = 0.00245 Wb or 2.45 * 10^{-3} Wb

(b) If the magnetic flux through the coil vanishes, the magnetic flux due to the coil will be:

Ф = LI

where L = inductance in the coil and I = current in the coil

Therefore:

L = Ф/I

L = 0.00245/3.8

L = 0.000645 H or 6.45 * 10^{-4} H

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A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

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3 years ago
What does the slope of a distance-time graph indicate?
lawyer [7]

Answer: It indicates the speed of a object. The steeper the line the greater the speed of the object.

8 0
2 years ago
Read 2 more answers
Suppose a cart with no fans has a starting velocity of 2 m/s. What will be the velocity of the cart when it reaches the wall?
Roman55 [17]

Answer:

less than stating velocity due to friction and air resistance.

Explanation:

3 0
2 years ago
A pendulum has 576 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
Natalka [10]
Ideally, 576 J  because energy is conserved.
In the real world, a tiny tiny tiny tiny bit less than 576 J ,
because we live in a world with friction and air resistance.
3 0
3 years ago
Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are
Sergio039 [100]

Answer:

21678.47223\ lbf-in^2

383.1109\ lbf-in^2

Explanation:

d = Diameter of column = 0.5 inch

A_c = Area of concrete = 119.4\ in^2

The strain in the system is conserved

\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s

Now

F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf

F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf

Stress is given by

\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2

The stress in the steel is 21678.47223\ lbf-in^2

\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2

The stress in the steel is 383.1109\ lbf-in^2

4 0
3 years ago
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