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2 years ago

12

If you want to maximize the magnetic force on a current in a conductor, how should you orient the current relative to the magnet

ic field?

1 answer:

tia_tia [17]2 years ago

7 0

Answer:

Should place the current perpendicular to the magnetic field

Explanation:

The magnetic force exerted on a current-carrying wire is given by

$F=ILBsin \theta$

where

I is the current in the wire

L is the length of the wire

B is the magnetic field

$\theta$ is the angle between the direction of the wire and the magnetic field

As we see from the formula, the magnetic force is maximum when

$sin \theta=1$

which means

$\theta=90^{\circ}$

So, when the current in the wire and the magnetic field are perpendicular to each other.

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A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy

SashulF [63]

Answer:

$5.444\times 10^{-4}$

Explanation:

The momentum of the neutron before and after the decay is the same since there's no external force.

$P_{sys}=const\\\\P=mv\\\\K=0.5mv^2$

#The neutron is initially at rest, so after the decay:

$P_A+P_B=0\\\\P_A=-P_B$

#After decay, the proton has +ve direction with a velocity $v_A$ while the electron moves in a negative direction with a velocity $v_B$

Therefore:

$P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B$

Let the energy released during the decay be Q:

$Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}$

Hence,Kp/Ktot is 5.444x10^(-4)

4 1

2 years ago

A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to

White raven [17]

Answer:

<h2>The potential difference increases </h2>

Explanation:

from the relation $E= \frac{V}{d}$

where E= electric field (force per coulomb)

V= voltage

d= distance

Hence the voltage is going to be V= E×d.

Therefore this means that increasing the distance increases the voltage.

3 0

2 years ago

A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r

Firlakuza [10]

Answer:

Speed of the this part is given as

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

$P_1 + P_2 + P_3 = P_i$

here we know that

$P_1 = P_2$

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

$P_1 + P_2 = \sqrt{P^2 + P^2}$

$P_1 + P_2 = \sqrt2 mv$

now from above equation we have

$P_3 = -(P_1 + P_2)$

$mv_3 = -(\sqrt 2 mv)$

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0

2 years ago

To prevent collisions and violations at intersections that have traffic signals, use the _____ to ensure the intersection is cle

Ann [662]

The answer should be:
<span>To prevent collisions and violations at intersections that have traffic signals, use the d</span>elayed acceleration technique<span> to ensure the intersection is clear before you enter it.
Delayed acceleration technique refers to w</span><span>aiting to go through an intersection until you have a chance to scan for other vehicles.</span>

3 0

2 years ago

7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has

NNADVOKAT [17]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, $\dfrac{m}{t}=30\ kg/s$

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

$F=\dfrac{\Delta P}{\Delta t}$

$F=\dfrac{m(v-u)}{\Delta t}$

$F=30\ kg/s\times (-16-16)\ m/s$

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

6 0

2 years ago