PHYSICS COLLEGE

A particle (q = 5.0 nC, m = 3.0 μg) moves in a region where the magnetic field has components Bx = 2.0 mT, By = 3.0 mT, and Bz = −4.0 mT. At an instant when the speed of the particle is 5.0 km/s and the direction of its velocity is 120° relative to the magnetic field, what is the magnitude of the acceleration of the particle in m/s2?

Answers

Answer 1
Answer:

The acceleration of the particle is 38.87 kg.

Net magnetic field

The net magnetic field is calculated as follows;

B_(net) = √(B_x^2 + B_y^2 + B_z^2) \n\nB_(net) = √(2^2 + 3^2 + 4^2) = 5.385 \ mT

Magnetic force on the charge

The magnetic force on the charge is calculated as follows;

F = qvB * sin(\theta)\n\nF = 5* 10^(-9) * 5* 10^3 * 5.385 * 10^(-3) * sin(120)\n\nF = 1.166 * 10^(-7) \ N

Acceleration of the particle

The acceleration of the particle is calculated as follows;

a = (F)/(m) \n\na = (1.166 * 10^(-7))/(3 * 10^(-9)) \n\na = 38.87 \ kg

Learn more about magnetic force here: brainly.com/question/13277365
Answer 2
Answer:

Explanation:

It is given that,

Charge on the particle, q=5\ nC=5* 10^(-9)\ C

Mass of the particle, m=3\ \mu g=3* 10^(-6)\ g=3* 10^(-9)\ kg

Magnetic field component, B_x=2\ mT,B_y=3\ mT,B_z=-4\ mT

Net magnetic field, B=√(2^2+3^2+4^2)=5.38\ mT=0.00538\ T

Speed of the particle, v = 5 km/s = 5000 m/s

Angle between velocity and magnetic field, \theta=120

Magnetic force is given by :

F=qvB\ sin\theta

F=5* 10^(-9)* 5000\ m/s* 0.00538* sin(120)

F=1.16* 10^(-7)\ N

Acceleration of the particle is given by, a=(F)/(m)

a=(1.16* 10^(-7)\ N)/(3* 10^(-9)\ kg)

a=38.6\ m/s^2

So, the acceleration of the particle is 38.6 m/s². Hence, this is the required solution.

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The speed of light in a transparent plastic is 2.5x108 m/s. If a ray of light in air(with n-1) strikes this plastic at an angle of incidence of 31.3 degrees, find the angle of the transmitted ray. Select one: a. 31.30 degrees b. 26.08 degrees c, 37.56 degrees d. 38.57 degrees e. 0.67 degrees

Answers

Answer:

Angle of transmitted ray is 29.14^(o)

Explanation:

According to snell's law we have

n_(1)sin(\theta _(i))=n_(2)sin(\theta r)

Since the incident medium is air thus we have n_(1)=1

By definition of refractive index we have

n=(c)/(v)

c = speed of light in vacuum

v = speed of light in medium

Applying values we get

n_(2)=(3* 10^(8))/(2.5* 10^(8))=1.2

Thus using the calculated values in Snell's law we obtain

sin(\theta _(r))=(sin(31.3)/(1.2)\n\n\therefore sin(\theta _(r))=0.4329\n\n\theta_(r)=sin^(-1)(0.4329)\n\n\theta _(r)=29.14

Answer:

Angle made by the transmitted ray = 25.65°

Explanation:

Speed of light in plastic = v = 2.5 × 10⁸ m/s  

refractive index of plastic (n₂) / refractive index of air (n₁)

= speed of light in air c / speed of light in plastic v.

⇒ n₂ = (3× 10⁸) / (2.5 × 10⁸) = 1.2

Angle of incidence = 31.3° = i  

n₁ sin i = n₂ sin r

⇒  sin r =  (1)(0.5195) / 1.2 = 0.4329

⇒ Angle made by the transmitted ray = r = sin⁻¹ (0.4329) =25.65°

             

Which of the following is a TRUE statement? a. It is possible for heat to flow spontaneously from a hot body to a cold one or from a cold one to a hot one, depending on whether or not the process is reversible or irreversible.
b. It is not possible to convert work entirely into heat.
c. The second law of thermodynamics is a consequence of the first law of thermodynamics.
d. It is impossible to transfer heat from a cooler to a hotter body.
e. All of these statements are false.

Answers

Answer:

e. All of these statements are false.

Explanation:

As we know that heat transfer take place from high temperature to low temperature.

It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.

The first law of thermodynamics is the energy conservation law.

Second law of thermodynamics  states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.

By using heat pump ,heat can transfer from cooler body to the hotter body.

Therefore all the answer is False.

(a) How fast and in what direction must galaxy A be moving if an absorption line found at 550 nm (green) for a stationary galaxy is shifted to 450 nm (blue) for A? (b) How fast and in what direction is galaxy B moving if it shows the same line shifted to 700 nm (red)?

Answers

Explanation:

For Part (a)

Since the apparent wavelength decreases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

=(v)/(c)\n v=(550-450)/(550)*3*10^(8)\n v=5.4545*10^(7)m/s

For Part (b)

Since the apparent wavelength increases hence galaxy moving towards the stationary observer.

Δλ/λ=v/c

=(v)/(c)\n v=(700-550)/(550)*3*10^(8)\n v=8.1818*10^(7)m/s

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade?

Answers

Incomplete question.The complete question is here

What is the magnitude of the force needed to hold the outer 2 cm of the blade to the inner portion of the blade? The outer edge of the blade is 21 cm from the center of the blade, and the mass of the outer portion is 7.7 g. Even though the blade is 21cm long, the last 2cm should be treated as if they were at a point 20cm from the center of rotation.

Answer:

F= 0.034 N

Explanation:

Given Data

Outer=2 cm

Edge of blade=21 cm

Mass=7.7 g

Length of blade=21 cm

The last 2cm is treated as if they were at a point 20cm from the center of rotation

To Find

Force=?

Solution

Convert the given frequency to angular frequency

ω = 45 rpm * (2*pi rad / 1 rev) * (1 min / 60 s)

ω= 3/2*π rad/sec

Now to find centripetal force.

F = m×v²/r

F= m×ω²×r

Put the data

F = 0.0077 kg × (3/2×π rad/sec)²× 0.20 m

F= 0.034 N

*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is the
acceleration of the ball?

Answers

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________. A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct

Answers

Answer:

4.25m/s

E. None of the option is correct

Explanation:

Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

Mathematically.

mu + MU = (m+M)v

m and M are the masses of the bullet and the block respectively

u and U are their respective velocities

v is their common velocity

from the question, the following parameters are given;

m = 20g = 0.02kg

u = 960m/s

M = 4.5kg

U =0m/s (block is at rest)

Substituting this values into the formula above to get v;

0.02(960)+4.5(0) = (0.02+4.5)v

19.2+0 = 4.52v

4.52v = 19.2

Dividing both sides by 4.52

4.52v/4.52 = 19.2/4.52

v = 4.25m/s

Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s
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