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Korvikt [17]
2 years ago
8

A single loop of wire with an area of 0.0900 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic

ular to the plane of the loop, and is decreasing at a constant rate of 0.190 T s. (a) What emf is induced in this loop? (b) If the loop has a resistance of find the current induced in the loop
Physics
1 answer:
erica [24]2 years ago
5 0

Answer:

(a) 0.0171 V

Explanation:

A = 0.09 m^2, dB/dt = 0.190 T/s

(a) According to the law of electromagntic induction

e = dФ / dt

e = A dB / dt

e = 0.09 x 0.190 = 0.0171 V

(b)

as we know

i = e / R

we can find induced current by dividing induced emf by resistance

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The first PCS technology used a form of time division multiplexing called ____.​ a. ​Time Division Multiple Access (TDMA) b. ​Gl
salantis [7]

Answer:

A. Time Division multiple Access (TDMA)

Explanation:

Time-division multiple access (TDMA) is a channel access method for shared-medium networks

It allows several users to share the same frequency channel by dividing the signal into different time slots. The users transmit in rapid succession, one after the other, each using its own time slot.

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2 years ago
Look at the diagram below. What is the size of the overall force on the boat?
inysia [295]
Can i get a picture pls
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2 years ago
What is the best definition of vectors?
mariarad [96]

Answer:

Number value and direction

Explanation:

Vectors are quantities that can be identified by value and direction . Examples are velocity and acceleration

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3 years ago
A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

  • Mass of the first puck = m_1\ =\ 5\ kg
  • Mass of the second puck = m_2\ =\ 3\ kg
  • initial velocity of the first puck = u_1\ =\ 3\ m/s.
  • Initial velocity of the second puck = u_2\ =\ -1.5\ m/s.

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = v_2\ =\ 2.31\ m/s.

Let v_1 be the final velocity of first puck after the collision.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = 25\times 10^{-3}\ sec

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0
3 years ago
If the sign of work is negative,
aivan3 [116]

If the sign of work is negative, that means the force and the motion are in opposite directions.

Let's say you see something roll off of the shelf. You catch it, and you let it down slowly and gently.

Gravity exerted down-force on it and it moved down. Gravity did positive work on it.

YOU exerted UP-force on it and it moved down. YOU did negative work on it.

(Also, the falling object exerted down-force on your hand, and your hand moved down. The falling object did positive work on your hand ! Where did THAT energy come from ? It came from the potential energy that the object had while it was on the shelf. Your hand absorbed that energy on the way down, doing negative work. So the object didn't have any kinetic energy when it reached the floor, and it did NOT splinter the floor or shatter in smithereens. It had barely enough energy left to make a sound when it hit the floor.)

5 0
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