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3 years ago

The direction of the force on a current-carrying wire in a magnetic field is

Physics

2 answers:

Bess [88]3 years ago

6 0

Answer:

Explanation:

The forces acting on a conductor carrying current placed in a magnetic field is analysed using the Fleming's left hand rule.

The rule states that "If the fire finger, the middle finger and the thumb are held mutually perpendicular to one another in a magnetic field, the fore finger acts in the direction of the magnetic field, the middle finger acts on the direction of the current while the thumb acts in the direction of the force.

Based on the rule, it can be inferred this current carrying wire placed in the magnetic field acts perpendicular to the magnetic field and force acting on the wire.

Yuliya22 [10]3 years ago

4 0

Answer:

The force that a particle of charge q and velocity v experiences when entering an electromagnetic field is:

F = q*(E + vxB)

Where E is the electric field and B is the magnetic field, here we only care for the magnetic field.

From this, we can see that the magnetic force will be perpendicular to the velocity and the magnetic field, and knowing that in a cable the electrons flow along the wire, we have that the force must be perpendicular to the wire and also to the magnetic field in where the wire is.

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Why potential energy become equal to kinetic energy at height

Gennadij [26K]

Answer:

because potentil energy is redy to go but its bound up

And kinetic energy is in motion

Explanation:

7 0

3 years ago

Read 2 more answers

John wants a new video tape and a new shirt, but he only has $12.00. He buys the video tape and gets $0.52 change. What was his

mestny [16]

Answer:

You have the answer in your comments. I will be copying it so your question doesn't get deleted.

The answers is $0.58

$11.48

the video tape

the new shirt

7 0

3 years ago

An 888.0 kg elevator is moving downward with a velocity of 0.800 m/s. It decelerates uniformly and comes to a stop in a distance

bagirrra123 [75]

Answer:

The value of tension on the cable T = 1065.6 N

Explanation:

Mass = 888 kg

Initial velocity ( u )= 0.8 $\frac{m}{sec}$

Final velocity ( V ) = 0

Distance traveled before come to rest = 0.2667 m

Now use third law of motion $V^{2}$ = $u^{2}$ - 2 a s

Put all the values in above formula we get,

⇒ 0 = $0.8^{2}$ - 2 × a ×0.2667

⇒ a = 1.2 $\frac{m}{sec^{2} }$

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

T = 1065.6 N

This is the value of tension on the cable.

5 0

3 years ago

An elevator cable breaks when a 925-kg elevator is 28.5 m above the top of a huge spring Ak = 8.00 * 104 N????mB at the bottom o

rosijanka [135]

Answer:

a) = 258352.5J

b) = 23.63 m/s

c) = 1.8m

Explanation:

Data;

Mass = 925kg

Distance (s) = 28.5m

Force constant (k) = 8.0*10⁴ N/m

g = 9.8 m/s²

a) = work = force * distance

But force = mass * acceleration

Force = 925 * 9.8 = 9065N

Work = F * s = 9065 * 28.5 = 258352.5J

b) acceleration (a) = (v² - u²) / 2s

a = v² / 2s

v² = a * 2s

v² = 9.8 * (2 * 28.5)

v² = 9.8 * 57

v² = 558.6

v = √(558.6)

V = 23.63 m/s

C). The work stops when the work done to raise the spring equals the work done to stop it by the spring

W = ½kx²

258352.5 = ½ * 8.0*10⁴ * x²

(2 * 258352.5) = 8.0*10⁴x²

516705 = 8.0*10⁴x²

X² = 516705 / 8.0*10⁴

X² = 6.46

X = √(6.46)

X = 2.54m

The compression was about 2.54m

3 0

3 years ago

Read 2 more answers

slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa

xenn [34]

Answer:

Maximum height reached by the rocket is

$y_{max} = 308 m$

total time of the motion of rocket is given as

$T = 16.44 s$

Explanation:

Initial speed of the rocket is given as

$v_i = 50 m/s$

acceleration of the rocket is given as

$a = 2 m/s^2$

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.68 m/s$

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 55.68^2 = 2(-9.81)(y - 150)$

$y_{max} = 308 m$

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

$v_f - v_i = at$

$55.68 - 50 = 2 t$

$t_1 = 2.84 s$

2) time to reach ground from this height

$\Delta y = v_y t + \frac{1}{2}gt^2$

$-150 = 55.68 t - \frac{1}{2}(9.81) t^2$

$t_2 = 13.6 s$

so total time of the motion of rocket is given as

$T = 13.6 + 2.84 = 16.44 s$

3 0

3 years ago