When an external magnetic field is applied, what happens to the protons in a sample?A) All protons align with the field.
B) All protons align opposite to the field.
C) Some protons align with the field and some align opposite to it.
D) All protons assume a random orientation.

Answers

Answer 1

Answer:

On account of external magnetic field, the protons will align with the magnetic field. Hence, option (a) is correct.

The given problem is based on the concept of magnetic field. The region where the magnetic force is experienced is known as magnetic field. Generally, the protons are the charged entities carrying the positive polarity and are one of the major constituents of modern atomic structure.

The origin of magnetic field occurs due to charged particles present in a specific space. And the magnetic field is due to the flowing of liquid metal in the outer core of the planet generates electric currents.
In the condition when an external field is applied, the majority of protons align to the field because these protons tend to act like small magnets under the effect of this external field.

Thus, we can conclude that on account of external magnetic field, the protons will align with the field.

Learn more about the magnetic field here:

brainly.com/question/14848188

Answer 2

Answer:

Some protons align with the field and some align opposite to it.

Explanation:

Majority align to the field because these protons tend to act like small magnets under the effect of this external field

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On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s meets the runner in a head-on collision. If the two players stick together, a) what is their velocity immediately after collision? b) What is the kinetic energy of the system just before the collision and a moment after the collision?

Answers

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

A magnet of mass 0.20 kg is dropped from rest and falls vertically through a 35.0 cm copper tube. Eddy currents are induced, causing the copper to warm up. The speed of the magnet as it emerges from the tube is 1.10 m/s. How much heat energy is dissipated to the environment?

Answers

Answer:

0.58 J

Explanation:

We know that Total energy is conserved.

Initial Kinetic energy + Initial potential energy = final kinetic energy+ final potential energy + dissipated heat energy

Initial kinetic energy = 0 ( magnet is at rest initially)

Initial Potential energy = m g h = (0.20 kg)(9.81 m/s²)(0.35 m) = 0.69 J

Final kinetic energy = 0.5 m v² = 0.5 ×0.20 kg × 1.10 m/s = 0.11 J

Final potential energy = 0

∴ Dissipated heat energy = (0.69 -0.11) J = 0.58 J

A 372-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 17.6 m/s2 , and its maximum speed is 1.75 m/s. a)Determine the angular frequency. b)Determine the amplitude. c)Determine the spring constant.

Answers

Answer with Explanation:

We are given that

Mass , m=372 g= $(372)/(1000)=0.372 Kg$

1 kg=1000g

Maximum acceleration, a= $17.6 m/s^2$

Maximum speed ,v=1.75 m/s

a.We know that

Maximum acceleration, a= $A\omega^2$

Maximum speed, v= $\omega A$

$17.6=A\omega^2$

$1.75=A\omega$

$(17.6)/(1.75)=(A\omega^2)/(A\omega)=\omega$

Angular frequency, $\omega=10.06 rad/s$

b.Substitute the value of angular frequency

$1.75=A(10.06)$

$A=(1.75)/(10.06)=0.17 m$

Hence, the amplitude=0.17 m

c.Spring constant,k= $m\omega^2$

Using the formula

$k=0.372* (10.06)^2$

Hence, the spring constant,k=37.6 N/m

A train is traveling at 30.0 m/sm/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 HzHz. The speed of sound in air should be taken as 344 m/sm/s.A. What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?B. What frequency frecede is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and receding from it?

Answers

Answer

given,

speed of sound = 344 m/s

speed of train = 30 m/s

frequency emitted by the train = 262 Hz

Doppler's effect

$f_L = (v + v_L)/(v + v_s)\ f_S$

f_L is the frequency of listener

f_S is the frequency of the source of the sound

v is the speed of the sound

v_L is the speed of listener.

v_S is the speed of the source

a) Speed of the passenger in another train , v = 18 m/s

another train is moving in opposite direction and approaching

v_L is positive as the listener is moving forward.

v_S is negative at the source is moving toward the listener.

$f_L = (344 + 18)/(344 - 30)* 262$

$f_L = 302\ Hz$

b) Speed of the passenger in another train , v = 18 m/s

another train is moving in opposite direction and receding

v_L is negative as the listener is moving away from source.

v_S is positive at the source is moving away the listener.

$f_L = (344 - 18)/(344 + 30)* 262$

$f_L = 228.37\ Hz$

How to Measure the mass of a coin using a triple beam balance

Answers

Answer:

https://youtu.be/stW-C7F7QOg

A tightly wound solenoid is 15 cm long, has 350 turns, and carries a current of 4.0 A. If you ignore end effects, you will find that the value of app at the center of the solenoid when there is no core is approximately