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charle [14.2K]
3 years ago
13

A long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of

magnitude 4.0\times 10^{-5}~\text{T}4.0×10 ​−5 ​​ T that is directed vertically downward, what is the resultant magnitude of the magnetic field 20 cm above the wire?
Physics
1 answer:
serg [7]3 years ago
8 0

Answer:

Explanation:

vertical magnetic field B_v = 4 x 10⁻⁵ T.

Magnetic field due to horizontal current at point 20 cm above

= (μ₀/4π ) x (2i / R)

= 10⁻⁷ x 2 x 20/ 20 x 10⁻²

= 2 x 10⁻⁵ T

It will act  coming out of paper. Hence it will be normal to magnetic field given .

So resultant magnetic field

= √ (4² + 2²) x 10⁻⁵ T

= 4.47 X 10⁻⁵ T .

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Alex787 [66]

Answer:

The speed of the two carts after the collision is 10 m/s.

Explanation:

Hi there!

The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:

initial momentum = mA · vA1 + mB · vB1

final momentum = (mA + mB) · vAB2

Where:

mA = mass of cart A = 0.500 kg

vA1 = velocity of cart A before the collision = 100 m/s

mB = mass of cart B = 1.50 kg.

vB1 = velocity of cart B before the collision = - 20 m/s

vAB2 = velocity of the carts that move as a single object = unknown.

(notice that we have considered leftward as negative direction)

Since the momentum of system remains constant:

initial momentum = final momentum

mA · vA1 + mB · vB1 = (mA + mB) · vAB2

Solving for vAB2:

(mA · vA1 + mB · vB1) / (mA + mB) = vAB2

(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2

vAB2 = 10 m/s

The speed of the two carts after the collision is 10 m/s.

6 0
3 years ago
A 50 kilogram woman wearing a seatbelt is traveling in a car that is moving with a velocity of +10 meters per second. In an emer
ra1l [238]

Answer:

the answer is A.) -1 * 10^3[N]

Explanation:

The solution consists of two steps, the first step is using the following kinematic equation:

v=v_{i} +a*t\\where:\\v=final velocity [m/s]\\v_{i}=initial velocity [m/s]\\a=acceleration[m/^2]\\t=time[s]\\

The initial velocity is 10 [m/s], and the final velocity is zero because the car stops in 0.5[s].

Replacing:

0=10+a*(0.5)\\a=-20[m/s^2]

Now in the second part, we need to use the second law of Newton, this law relates the forces with the acceleration of a body.

In the moment when the car stops suddenly the driver will feel the force of the seatbelt acting in the opposite direction of the movement.

F=m*a\\F=50[kg]*(-20[m/s^2])\\units\[kg]*[m/s^2]=[N]\\F=-1000[N] or -1*10^{3} [N]

The minus sign means that the force is acting in the opposite direction of the movement.

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