PHYSICS COLLEGE

A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.20 107 m/s and experiences an acceleration of 1.90 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer 1

Answer:

Answer:

The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Explanation:

Given that,

Speed $v = 2.20*10^7\ m/s$

Acceleration $a=1.90*10^(13)\ m/s^2$

We need to calculate the magnetic field

Using formula of magnetic field

$F=qvB$ ....(I)

Using newton's second law

$F= ma$ ....(II)

From equation (I) and (II)

$ma=qvB$

Put the value into the formula

$1.90*10^(13)*1.67*10^(-27)=1.6*10^(-19)*2.20*10^(7)*B$

$3.173*10^(-14)=1.6*10^(-19)*2.20*10^(7)*B$

$B=(3.173*10^(-14))/(1.6*10^(-19)*2.20*10^(7))$

$B=0.009014\ T$

We need to calculate the direction of the field

Using the right hand rule, point the right hand fingers along the velocity which is in the positive z direction.

Now, if we curl the fingers along the direction of magnetic field that is in the negative y direction, then the thumb will point in the positive x direction.

Hence, The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Related Questions

ou are writing a short adventure story for your English class. In your story, two submarines need to arrive at a place in the middle of the Atlantic Ocean at the same time. They start out at the same time from positions equally distant from the rendezvouspoint. They travel at different speeds, but both go in a straight line. The first submarine travels at an average speed of 20 km/hr for the first 500 km, 40 km/hr for the next 500 km, 30 km/hr for the next 500 km and 50 km/hr for the final 500 km. In the story’s plot, the second submarine is required to travel at a constant velocity, so the captain needs to determine the magnitude of that velocity. What is that velocity?
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)
An unstable atomic nucleus of mass 1.82 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.18 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) Find the velocity of the third particle.
Find the intensity III of the sound waves produced by four 60-WW speakers as heard by the driver. Assume that the driver is located 1.0 mm from each of the two front speakers and 1.5 mm from each of the two rear speakers.
A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?

While leaning out a window that is 6.0 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.6 m above the ground. Determine the following.(a) the amount of work done by the force of gravity on the ball.(b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released.(c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught.

Answers

Answer:

a) W = 25.872 J

b) - 35.28 J

c) - 9.408

Explanation:

a) The amount of work done by the force of gravity on the ball = Change in potential energy between the two vertical points = - mg (H₂ - H₁)

F = - mg (gravity is acting downwards)

F = - 0.6 × 9.8 = - 5.88 N

(H₂ - H₁) = (1.6 - 6) = - 4.4 m

W = (-5.88)(-4.4) = 25.872 J

b) Gravitational-potential energy of the ball when it was released relative to the ground = (- mg) H₁ = (- 0.6 × 9.8) × 6 = - 35.28 J

c) Gravitational-potential energy of the ball when it is caught relative to the ground = (-mg)(H₂) = -0.6 × 9.8 × 1.6 = - 9.408 J

A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes. If instead of a single wire we use a coil of thin Nichrome wire containing 23 turns, what does the ammeter read?

Answers

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

$\epsilon=NA(dB)/(dt)$

For N = 1

$\epsilon=A(dB)/(dt)$

We need to calculate the current

Using formula of current

$i=(\epsilon)/(R)$

Put the value of emf

$i=(A(dB)/(dt))/(R)$

Now, if the number of turn is 22 , then induced emf would be

$\epsilon'=NA(dB)/(dt)$

Then the current would be

$i'=(\epsilon')/(NR)$

$i'=(NA(dB)/(dt))/(NR)$

$i'=(A(dB)/(dt))/(R)$

$i'=i$

Hence, The current would be same in both situation.

The table below shows the mass and velocity of four objects. Which object has the least inertia?A. Y
B. Z
C. X
D. W

Answers

The object with the least inertia is Z.

option B is the correct answer.

What is Newton's first law of motion?

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that path unless it is acted upon by an external force and it will move in the direction of applied force.

The Newton's first law of motion is also called the law of inertia because it depends on the mass of the object.

Inertia is defined as the reluctancy of an object to move when a force is applied to it.

As the mass of an object increase, the inertia of the object increases because the object will be more reluctant to move when a force is applied to it.

Thus, the more massive an object is, the greater the object's inertia and vice versa.

Learn more about inertia here: brainly.com/question/1140505

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A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel.What is the battery's internal resistance?

Answers

Answer:

0.46Ω

Explanation:

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

First, let's calculate the effective resistance in the circuit:

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = $(V^(2) )/(R)$

=> R = $(V^(2) )/(P)$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $(V^(2) )/(P)$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $(12.0^(2) )/(4)$

R₁ = $(144)/(4)$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $(V^(2) )/(P)$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $(12.0^(2) )/(4)$

R₂ = $(144)/(4)$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$(1)/(R_(X) )$ = $(1)/(R_1)$ + $(1)/(R_2)$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$(1)/(R_X)$ = $(1)/(36)$ + $(1)/(36)$

$(1)/(R_X)$ = $(2)/(36)$

Rₓ = $(36)/(2)$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

Now calculate the current I, flowing in the circuit:

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $(11.7)/(18)$

I = 0.65A

Now calculate the battery's internal resistance:

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $(0.3)/(0.65)$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

Answer:

$R_i_n_t=0.45 \Omega$

Explanation:

Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:

$R_i_n_t=((V_N_L)/(V_F_L) -1)R_L$

Where:

$V_F_L=Load\hspace{3}voltage=11.7V\nV_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\nR_L=Load\hspace{3}resistance$

As you can see, we don't know the exactly value of the $R_L$ . However we can calculated that value using the next simple operations:

The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:

$(12)/(11.7) =(4)/(P)$

Solving for $P$ :

$P=(11.7*4)/(12) =3.9W$

Now, the electric power is given by:

$P=(V^2)/(R_b)$

Where:

$R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb$

So:

$R_b=(V^2)/(P) =(11.7^2)/(3.9) =35.1\Omega$

Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:

$(1)/(R_L) =(1)/(R_b) +(1)/(R_b) =(2)/(R_b) \n\n R_L= (R_b)/(2) =(35.1)/(2)=17.55\Omega$

Finally, now we have all the data, let's replace it into the internal resistance equation:

$R_i_n_t=((12)/(11.7) -1)17.55=0.45\Omega$

If the rise and fall of your lungs is considered to be simple harmonic motion, how would you relate the period of the motion to your breathing rate (breaths per minute)? Breaths per minute is an angular frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is the square root of that value. Breaths per minute is a frequency. The period is its reciprocal. Breaths per minute is an angular frequency. The period is its reciprocal.

Answers

Answer:

Breaths per minute is a frequency. The period is its reciprocal.

Explanation:

In simple harmonic motion, a period (T) is the time taken for one point to start in a position and reach that position again, in other words to complete a cycle or lapse. In this case, a period is the time one takes from starting to inspire the air to releasing all of it from the lungs.

In simple harmonic motion, the frequency (f) is how many times a point completes a cycle or lapse in one unity of time (could be one second, one minute, one hour, etc). In this case, the frequency is how many times one breathes in one minute. This is the breathing rate, since it is breathings per minute. Breaths per minute is a frequency.

Period (T) and frequency (f) relate to each other in the following formulae: $T=(1)/(f)$ or $f=(1)/(T)$ .

Therefore, breaths per minute is a frequency, and since it is related to the period, we say the period is reciprocal to it.

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