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weeeeeb [17]
4 years ago
6

A straight wire lies along the y-axis initially carrying a current of 10 A in the positive y-direction. The current decreases an

d reverses to 10 A in the negative y-direction, the change in current happening at a uniform rate. In the 1st quadrant a square conducting coil has 2 sides parallel to the y-axis and the other 2 sides parallel to the x-axis. The side of the coil nearest and parallel to the straight wire is at a distance equal to the length of one of the sides of the square. As the current is going from 10 A in one direction to 10 A in the other, in which direction is the induced current in this side of the square coil nearest to the straight wire

Physics
1 answer:
Elan Coil [88]4 years ago
8 0

Answer:

Explanation:

The magnetic field due to straight wire is into the square coil.

As the current in straight wire decreases the magnetic flux in the coil decreases . The induced magnetic field is into the coil.The induced current is along +y direction

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A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm10nm. Assume that the walls act
Mademuasel [1]

Answer:

the correct answue are  B, A, C, C, B

Explanation:

1) The electric field is requested, let's approximate the membrane by a parallel plate with surface charge density

         E = \frac{\sigma }{2 \epsilon_o }

         E = \frac{ 10^{-5}}{2 \ 8.85 \ 10^{-12}}

         E = 5.65 10⁵ N / C

the correct answer is B

2) A calcium ion has two positive charges, so the force applied by each side of the membrane (plate)

         F = q E

         F = 2  1.6 10⁻¹⁹  5.65 10⁵

         F = 1.8 10⁻¹³ N

the total force is the sum of the force of each membrane and the two forces go to the same side

         F = total = 2 F

         F_total = 3.6 10⁻¹³ N

the correct answer is A

3) the field and the electric potential are related

          ΔV = - E s

          ΔV = - 5.65 10⁵  10 10⁻⁹

          ΔV = - 5.65 10⁻³ V

          the correct answer is C

4) In the exercise they indicate that the outer wall has a positive charge, therefore, as they indicate that we approximate the system to a capacitor, the inner wall must be negatively charged.

The electric field goes from the positive to the negative charge, which is why it goes from the outer wall to the inner wall

the correct answer is C

5) For this part we use conservation of energy

starting point. On the inside wall, brown

            Em₀ = U = qV

final point. On the outside

             Em_f = K

energy is conserved

           Em₀ = Em_f

           q V = K

            K = 3 10⁻¹⁵  5.65 10⁻³

            K = 1.7 10⁻¹⁷ J

the correct answer is B

4 0
3 years ago
Give main forms before and after for the following energy transformations: A gasoline engine.
klemol [59]
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ It would be chemical energy to kinetic energy. The chemical energy from the gasoline is being transferred to kinetic energy.

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

4 0
4 years ago
An infinite slab of charge of thickness 2z0 lies in the xy-plane between z=−z0 and z=+z0. The volume charge density rho(C/m3) is
Oksi-84 [34.3K]

Answer:

please read the answer below

Explanation:

To find the electric field you can consider the Gaussian law for a cylindrical surface inside the slab.

\int E dA=EA_{G}=\frac{Q_{int}}{\epsilon_o}

Q_{int}=\rho V_{G}

where Qint is the charge inside the Gaussian surface, AG is the area of the surface and rho is the charge density of the slab.

By using the formula for the volume of a cylinder you obtain:

V_{G}=\pi r^2h

where h is the height. If you assume that the slab is in the interval (-zo<z<z0) you can write VG:

V_{G}=\pi r^2 z

Finally, by replacing in the expression for E you get:

E=\frac{Q_{int}}{\epsilon_o A_G}=\frac{Q_{int}}{\epsilon_o \pi r^2}\frac{z}{z}=\frac{\rho z}{\epsilon_o}

E=\rho z/\epsilon_o

hence, for z>0 you obtain E=pz/eo > 0

for z<0 -> E=pz/eo < 0

7 0
3 years ago
A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static fr
PSYCHO15rus [73]

Answer:

ω = 2.1 rad/sec

Explanation:

  • As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
  • This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
  • Now, we need to ask ourselves: what supplies this force?
  • In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
  • We know that this force can be expressed as follows:

       f_{frs} = \mu_{s} * F_{n} (1)

      where μs = coefficient of static friction between the rock and the merry-

      go-round surface = 0.7, and Fn = normal force.

  • In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
  • Fn = m*g (2)
  • This static friction force is just the same as the centripetal force.
  • The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

       F_{c} = m* \omega^{2}*r (3)

  • Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

       \omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec

  • This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.
4 0
3 years ago
The fact that the total amount of energy in a system remains constant is a(n) experiment. theory. hypothesis. Or law.
Murrr4er [49]

Answer:

<h3>law</h3>

Explanation:

The law of conservation of energy states that the total amount of energy in a system remains constant.

8 0
3 years ago
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