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The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horizontally from north to south. At the electron's latitude the vertical component of the Earth's magnetic field points down with a magnitude of 42.3 μT. What is the direction of the force on the electrons due to this component of the magnetic field?

Answers

Answer 1

Answer:

Answer:

The direction is due south

Explanation:

From the question we are told that

The energy of the electron is $E = 19.0keV = 19.0 *10^3 eV$

The earths magnetic field is $B = 42.3 \muT = 42.3 *10^(-6) T$

Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

$\= F = q (\= v * \=B)$

On the first uploaded image is an illustration of the movement of the electron

Looking at the diagram we can see that in terms of direction the magnetic force is

$\= F =q(\=v * \= B)= -( -\r i * - \r k)$

$= -(- (\r i * \r k))$

generally i cross k = -j

so the equation above becomes

$\= F = -(-(- \r j))$

$= - \r j$

This show that the direction is towards the south

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A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very far from the wire. What is the ratio Ff/Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?

Answers

The ratio of the electric force on the proton after the wire segment is shrunk to three times its original length to the force before the segment was shrunk is 3.

The electric force between a point charge and a segment of wire with a distributed charge is given by Coulomb's law.

The formula for the electric force on a point charge q due to a segment of wire with charge Q distributed along its length L is:

$F=(k.q.Q)/(L)$

where:

F is the electric force on the point charge,

k is Coulomb's constant ( 8.988 × 1 0⁹ Nm²/ C²),

q is the charge of the point charge,

Q is the charge distributed along the wire segment, and

L is the length of the wire segment.

When the wire segment is shrunk to one-third of its original length, the new length becomes 1/3 L.

The chargedistribution remains the same, only the length changes.

So, the new electric force $F_f$ on the proton after the segment is shrunk becomes:

$F_f=(k.q.Q)/((1)/(3)L)$

The original electric force $F_i$ on the proton before the segment was shrunk is:

$F_i = (k.q.Q)/(L)$

let's find the ratio $(F_f)/(F_i)$ :

$(F_f)/(F_i)=((k.q.Q)/((1)/(3)L))/((k.q.Q)/(L))$

$(F_f)/(F_i)=3$

Hence, the ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk is 3.

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Final answer:

The ratio of the electric force on the proton after the wire segment is shrunk is equal to the ratio of their charges.

Explanation:

The ratio of the electric force on the proton after the wire segment is shrunk to the force before the segment was shrunk can be found using Coulomb's law. Coulomb's law states that the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

In this case, the charges involved are the charge of the wire segment and the charge of the proton. Since the wire segment contains 10 nC of charge, we can consider it as one of the charged objects. The proton is very far from the wire, so we can assume that the distance between them remains the same before and after the wire segment is shrunk. Therefore, the ratio of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk is equal to the ratio of their charges.

Let's assume that the initial force on the proton is Fi and the final force on the proton is Ff. Using the given information, we have:

Fi = k(q1 * q2) / r^2

where k is the electrostatic constant, q1 and q2 are the charges of the wire segment and the proton respectively, and r is the distance between them.

After the wire segment is shrunk to one-third of its original length, the charge of the wire segment remains the same and the distance between the wire segment and the proton also remains the same. Therefore, the ratio Ff/Fi can be calculated as:

Ff/Fi = (q1 * q2) / (q1 * q2) = 1

Can a car moving with a negative velocity moves faster than a car moving with a positive velocity? explain.

Answers

Answer:

Yes.

Explanation:

This is because "negative velocity" just means it is in the negative in relation to the point of 0. Negative velocity doesn't equal a decrease in velocity. For example lets say you were parked next to a cone (this cone represents zero) if you accelerate forwards then that would be positive acceleration. If you were to accelerate backwards, this would be in the negative direction, aka negative velocity.

SUMMARY:

A negative velocity means that the object which has the negative velocity is moving in the opposite direction of an object moving at a positive velocity. This is a question of frame of reference. The possibility for the velocity is what makes it different to the speed. Speed is only positive.

Final answer:

In physics, a negative velocity can be faster than a positive one when considering speed alone, but not when considering motion direction. For instance, if a car is moving faster but in an opposite direction, it will have a higher speed but a negative velocity.

Explanation:

In physics, the term velocity represents both the speed of an object and its direction of motion. A negative velocity simply means that the object is moving in the opposite direction of the reference point. So, a car moving with a negative velocity can 'move faster' than a car moving with positive velocity if you're considering its speed alone.

Let's assume you have Car A moving at a speed of 40 km/hr in the eastern direction (positive velocity) and Car B moving at a speed of 60 km/hr in the western direction (negative velocity). Even though Car B is described as having a negative velocity, it is moving faster than Car A in terms of speed.

However, remember that in physics, direction matters when considering velocity. So, if you compare their velocities without ignoring the direction, Car A is moving faster to the east than Car B is to the west, even if Car B has higher speed.

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A runner first runs a displacement A of 3.20 km due south, and then a second displacement B that points due east. (a) The magnitude of the resultant displacement A + B is 5.38 km. What is the magnitude (in m) of B?

Answers

Answer: 4,438.96m

Explanation:

(kindly find attachment below)

From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.

By using phythagoras theorem

H² = O² + A²

(5.38)² = (3.20)² + B²

28.944 = 10.24 + B²

B² = 28.944 - 10.24

B² = 18.7044

B = √18.7044

B = 4.439km to meter is 4.439 * 1000 = 4,438. 96m

Answer:

B = 4325 m

Explanation:

Resolving the displacement into x and y components.

Let north = positive y component

East = positive x component

So,

Rx = B

Ry = -3.20 km

Magnitude of the resultant displacement is

R = √(B^2 + (-3.20)^2)

R is given as R = 5.38 km

Making B the subject of formula;

B = √(R^2 - (-3.20)^2)

B = √(5.38^2 - (-3.20)^2)

B = 4.325 km

B = 4325 m

The knot at the junction is in equilibrium under the influence of four forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 5.7 N force acts from above on the right at an angle of 50◦ with the horizontal. The 6.2 N force acts from below on the right at an angle of 44◦ with the horizontal. The 6.7 N force acts from below on the left at an angle of 43◦ with the horizontal.1. What is the magnitude of the force F?
2. What is the angle a of the force F in the figure above?

Answers

(a) The magnitude of the force F acting on the knot is 5.54 N.

(b) The angle α of the force F is 54.4⁰.

The given parameters:

F force at α
5.7 N force at 50⁰
6.2 N force at 44⁰
6.7 N force at 43⁰

The net vertical force on the knot is calculated as follows;

$F_y = Fsin(\alpha) + 5.7 sin(50) - 6.2 sin(44) - 6.7 sin(43)\n\nF_y = F sin(\alpha) -4.51\n\nFsin(\alpha) = 4.51$

The net horizontal force on the knot is calculated as follows;

$F_x = -F cos(\alpha) + 5.7 cos(50) + 6.2cos(44) - 6.7cos(43)\n\nF_x = -Fcos(\alpha) + 3.22\n\nFcos(\alpha) = 3.22$

From the trig identity;

$sin^2 \theta + cos^ 2 \theta = 1\n\n$

$(Fsin(\alpha))^2 + (Fcos(\alpha))^2 = (4.51)^2 + (3.22)^2\n\nF^2(sin^ 2\alpha + cos^2 \alpha) = 30.71\n\nF^2(1) = 30.71\n\nF = √(30.71) \n\nF = 5.54 \ N$

The angle α of the force F is calculated as follows;

$Fsin(\alpha) = 4.51\n\nsin(\alpha) = (4.51)/(F) \n\nsin(\alpha ) = (4.51)/(5.54) \n\nsin(\alpha ) = 0.814\n\n\alpha = sin^(-1)(0.814)\n\n\alpha = 54.5 \ ^0$

Find the image uploaded for the complete question.

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The knot is in equilbrium, so there is no net force acting on it. Starting with the unknown force and going clockwise, denote each force by F₁, F₂, F₃, and F₄, respectively. We have

F₁ + F₂ + F₃ + F₄ = 0

Decomposing each force into horizontal and vertical components, we have

F cos(180º - α) + (5.7 N) cos(50º) + (6.2 N) cos(-44º) + (6.7 N) cos(-137º) = 0

F sin(180º - α) + (5.7 N) sin(50º) + (6.2 N) sin(-44º) + (6.7 N) sin(-137º) = 0

Recall that cos(180º - x) = - cos(x) and sin(180º - x) = sin(x), so these equations reduce to

F cos(α) ≈ - 3.22 N

F sin(α) ≈ 4.51 N

(1) Recall that for all x, sin²(x) + cos²(x) = 1. Use this identity to solve for F :

(F cos(α))² + (F sin(α))² = F ² ≈ 30.73 N² → F ≈ 5.5 N

(2) Use the definition of tangent to solve for α :

tan(α) = sin(α) / cos(α) ≈ 1.399 → α ≈ 126º

or about 54º from the horizontal from above on the left of the knot.

A 39 kg block of ice slides down a frictionless incline 2.8 m along the diagonal and 0.74 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?

Answers

Answer:

(a) Fw = 101.01 N

(b) W = 282.82 J

(d) N = 368.61 N

(e) Net force = 0 N

Explanation:

(a) In order to calculate the magnitude of the worker's force, you take into account that if the ice block slides down with a constant speed, the sum of forces, gravitational force and work's force, must be equal to zero, as follow:

$F_g-F_w=0$ (1)

Fg: gravitational force over the object

Fw: worker's force

However, in an incline you have that the gravitational force on the object, due to its weight, is given by:

$F_g=Wsin\theta=Mg sin\theta$ (2)

M: mass of the ice block = 39 kg

g: gravitational constant = 9.8m/s^2

θ: angle of the incline

You calculate the angle by using the information about the distance of the incline and its height, as follow:

$sin\theta=(0.74m)/(2.8m)=0.264\n\n\theta=sin^(-1)(0.264)=15.32\°$

Finally, you solve the equation (1) for Fw and replace the values of all parameters:

$F_w=F_g=Mgsin\theta\n\nF_w=(39kg)(9.8m/s^2)sin(15.32\°)=101.01N$

The worker's force is 101.01N

(b) The work done by the worker is given by:

$W=F_wd=(101.01N)(2.8m)=282.82J$

(c) The gravitational force on the block is, without taking into account the rotated system for the incline, only the weight of the ice block:

$F_g=Mg=(39kg)(9.8m/s^2)=382.2N$

The gravitational force is 382.2N

(d) The normal force is:

$N=Mgcos\theta=(39kg)(9.8m/s^2)cos(15.32\°)=368.61N$

(e) The speed of the block when it slides down the incle is constant, then, by the Newton second law you can conclude that the net force is zero.

A baseball player throws a ball into the stands at 15.0 m/s and at an angle 45.0° above the horizontal. On its way down, the ball is caught by a spectator 4.10 m above the point where the ball was thrown. How much time did it take for the ball to reach the fan in the stands?

Answers

Answer:

Time = 1.61 seconds

Explanation:

Using the equation displacement of a trajectory motion in the y plane

Y = u t sin ů - ½gt²....equation 1 where

Y= vertical displacement =4.1

U = initial velocity = 15m/s

g = acc. Due to gravity = 10m/s

Ů = angle of trajectory = 45

t = time to reach fan on its way down

Sub into equ 1

4.1 = 15t sinů - ½ * 10t²

4.1 = 10.61t - 5t²

Solve using quadratic formula

t =[-B±( -B² -4AC)^½]/2A....equation 2

Where A = 5, B=10.61, C =4.1

Substitute A,B,C into equ2

t = (10.61±5.53)/10

t = 0.508seconds or 1.61seconds

Since it is on its way down t= 1.61 seconds

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