A moderate wind accelerates a pebble over a horizontal xy plane with a constant acceleration a with arrow = (4.60 m/s2)i hat + (7.00 m/s2)j. At time t = 0, the velocity is (4.3 m/s)i hat. What are magnitude and angle of its velocity when it has been displaced by 11.0 m parallel to the x axis?

A Moderate Wind Accelerates A Pebble Over A Horizontal Xy Plane With A Constant Acceleration A With Arrow

Answers

Answer 1

Explanation:

Given

Acceleration of the pebble is

At t=0, velocity is

considering horizontal motion

[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]

Velocity acquired during this time

[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]

Consider vertical motion

[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]

Net velocity is

[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]

Angle made is

[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]


Related Questions

A child with a weight of 230 N swings on a playground swing attached to 2.20-m-long chains. What is the gravitational potential energy of the child-Earth system relative to the child's lowest position at the following times?
(a) when the chains are horizontal (in J)
(b) when the chains make an angle of 33.0° with respect to the vertical (in J)
(c) when the child is at his lowest position (in J)

Answers

Answer:

a)  U = 506 J, b)  U = 37.11 J, c) U = 0

Explanation:

The gravitational power energy is given by the expression

         U = m g (y -y₀)

In general, a reference system is set that allows the expression to be simplified, in this case let's assume the reference system at the child's lowest point, therefore y₀ = 0

Let's use trigonometry to find the child's height

          h = y = L - L cos θ

         

we substitute

           U = m g L (1 - cos θ)

a) when the chain is horizontal θ = 90 and cos 90 = 0

           U = mg L

weight and mass are related

            W = mg

            m = W / g

           

           

           U = 230 2.20

           U = 506 J

b) θ = 33.0º

           cos 33 = 0.83867

           U = 230 (1 - 0.83867)

           U = 37.11 J

c) in this case θ = 0 cos 0 = 1

            U = 0

a. Give an example of the conversion of light energy to electrical energy.

b. Give an example of chemical energy converting to heat energy.

c. Give an example of mechanical energy converting to heat energy.

Answers

Explanation:

a) photovoltaic cell is a semiconductor device and it converts light energy to electrical energy

b) burning of coal converts chemical energy to heat energy

c) rubbing of both hands against each other converts mechanical to heat energy

Answer:

a. solar cells

b.coal,wood,petroleum

c.rubbing ours palms

You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet

Answers

6 cubic feet I’m pretty sure that’s the answer

trình bày nguyên lý Đa lăm be

Answers

I don’t understand this question

A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positive direction, what is the truck's resultant displacement

Answers

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

A ball has a mass of 4.65 kg and approximates a ping pong ball of mass 0.060 kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00 m / s, determine the final velocities of both masses after the collision. Use equations 9.21 and 9.22 from the textbook. The book is on WebAssign.

Answers

Answer:

the final velocity of the ball is 4.87 m/s

the final velocity of the ping ball is 9.87 m/s

Explanation:

Given;

mass of the ball, m₁ = 4.65 kg

mass of the ping ball, m₂ = 0.06 kg

initial velocity of the ping ball, u₂ = 0

initial velocity of the ball, u₁ = 5 m/s

let the final velocity of the ball = v₁

let the final velocity of the ping ball, = v₂

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁  +  m₂u₂  = m₁v₁   +  m₂v₂

4.65(5)  +   0.06(0)   =   4.65v₁   +   0.06v₂

23.25 + 0 = 4.65v₁  +  0.06v₂

23.25 = 4.65v₁  +  0.06v₂  ------ (1)

Apply one-directional velocity equation;

u₁ + v₁ = u₂  +  v₂

5 + v₁ = 0  +  v₂

5 + v₁ = v₂

v₁ = v₂ - 5  -------- (2)

substitute equation (2) into (1)

23.25 = 4.65(v₂ - 5)  +  0.06v₂

23.25 = 4.65v₂  -  23.25   +   0.06v₂

46.5 = 4.71 v₂

v₂ = 46.5/4.71

v₂ = 9.87 m/s

v₁ = v₂ - 5

v₁ = 9.87 - 5

v₁ = 4.87 m/s

what is conservation energy?

Answers

Explanation:

Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant

hope it is helpful to you

A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill?
A) 3.57 m.
B) 4.28 m.
C) 3.14 m.
D) 2.68 m.

Answers

Answer:

A(3.56m)

Explanation:

We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.

We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.

Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2

Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2=  v^2).

Ef= -mgh+3/4mv^2

Since Ef=Ei=0

Mgh=3/4mv^2

gh=3/4v^2

h=0.75v^2/g

plug in givens to get h= 3.57m

A piston-cylinder device contains 5 kg of refrigerant-134a at 0.7 MPa and 60°C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 24°C. If the surroundings are at 100 kPa and-24°C, determine:
(a) the exergy of the refrigerant at the initial and the final states and
(b) the exergy destroyed during this process.

Answers

Answer:

Yes sure, keep it going, and never give up because your dreams are so important

A) The exergy of the refrigerant at the initial and final states are :

Initial state =  - 135.5285 kJ Final state =  -51.96 kJ

B) The exergy destroyed during this process is : - 1048.4397 kJ

Given data :

Mass ( M )  = 5 kg

P1 = 0.7 Mpa = P2

T1 = 60°C = 333 k

To = 24°C = 297 k

P2 = 100 kPa

A) Determine the exergy at initial and final states

At initial state :

U = 274.01 kJ/Kg , V = 0.034875 m³/kg , S = 1.0256 KJ/kg.k

exergy ( Ф ) at initial state = M ( U + P₂V - T₀S )

                                           = 5 ( 274.01 + 100* 10³ * 0.034875 - 297 * 1.0256)

                                           ≈ - 135.5285 kJ

At final state  :

U = 84.44 kJ / kg , V = 0.0008261 m³/kg,  S = 0.31958 kJ/kg.k

exergy ( ( Ф ) at final state = M ( U + P₂V - T₀S )

                                             = -51.96 kJ

B) Determine the exergy destroyed

  exergy destroyed = To * M ( S2 - S1 )

                                 = 297 * 5 ( 0.31958 - 1.0256 )

                                 = - 1048.4397 KJ

Hence we can conclude that A) The exergy of the refrigerant at the initial and final states are : Initial state =  - 135.5285 kJ, Final state =  -51.96 kJ  and The exergy destroyed during this process is : - 1048.4397 kJ

Learn more about exergy : https://brainly.com/question/25534266

Riley, a student, notices that the protractor tool does not measure the angle just as the ball leaves the surface. She sees that the ball must travel some distance before it crosses the protractor, so the direction of travel may have changed as the ball moves upwards. She says that this is the cause of the discrepancy between her predicted angle and the measured angle. Does this reasoning explain the discrepancy between your predicted angle and your measured angle. Use evidence to support your claim.

Answers

Answer:

Riley's reasoning is correct

Explanation:

Her reasoning is correct because as the ball moves upwards, the acceleration due to gravity would be vertical and in downward position. Therefore at all points as the ball moves, the velocity of the ball is going to change in magnitude as well as in direction. given that the direction keeps changing at certain points, the angle made by the initial velocity just as the ball left the surface would also have to continuously change.

If Riley has to wait for this ball to move some inches before she uses the protractor to measure the angle, the angle of travel would have to change.

Therefore there is going to be discrepancies between the measured angle and the predicted angle. The predicted is the angle of velocity with the horizontal just as this ball moves from the surface.

Many collisions, like the collision of a bat with a baseball, appear to be instantaneous. Most people also would not imagine the bat and ball as bending or being compressed during the collision. Consider the following possibilities: The collision is instantaneous. The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact. The collision takes a finite amount of time, during which the ball and bat are bending or being compressed. How can two of these be ruled out based on energy or momentum considerations?
The collision is instantaneous.
The collision takes a finite amount of time, during which the ball and bat retain their shapes and remain in contact.
The collision takes a finite amount of time, during which the ball and bat are bending or being compressed.
How can two of these be ruled out based on energy or momentum considerations?

Answers

Answer:

The collision takes a finite amount of time, during which the ball and bat are bending or being compressed

Explanation:

These two conditions can be ruled out on the fact that :The collision takes a finite amount of time, during which the ball and bat are bending or being compressed

The rule of energy is been broken here because during the collision of objects energy and momentum is conserved. i.e. the change in shape of the ball when hit by the bat should not be noticed because the compression and returning to normal shape happens instantaneously

Three 30 g metal balls, one of aluminum, copper and lead, are placed in a large beaker of hot water for a few minutes. [The specific heats of aluminum, copper, and lead are 903, 385, and 130 J / (kg ° C), respectively].
to. Which of the balls, if any, will reach the highest temperature? Explain.
b. Which of the balls, if any, will have the most heat energy? Explain.

Answers

Answer:

The answer is below

Explanation:

Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.

a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.

Hence lead will have the highest temperature since it has the least specific heat capacity.

b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.

Hence aluminum will have the most heat since it has the highest specific heat capacity.

As it pulls itself up to a branch, a chimpanzee accelerates upward at 2.4 m/s2 at the instant it exerts a 260-N force downward on the branch.Find the magnitude of the force the chimpanzee exerts on the Earth.

Answers

Answer:

[tex]F=208.83N[/tex]

Explanation:

From the question we are told that:

Acceleration [tex]a=2.4m/s^2[/tex]

Force of Branch [tex]F=260N[/tex]

Generally the Newton's equation second law for Force is mathematically given by

 [tex]ma=F-mg[/tex]

 [tex]m=\frac{260}{2.4+9.8}[/tex]

 [tex]m=21.31kg[/tex]

Therefore

 [tex]F=mg[/tex]

 [tex]F=(21.31)(9.8)[/tex]

 [tex]F=208.83N[/tex]

Physics is killing me. Any help?

Answers

Answer:

The simplified expression is  3.833 x 10⁷ g

Explanation:

Given expression;

3.88 x 10⁷ g  -  4.701 x 10⁵ g

The expression above is simplified as follows;

= (3.88 x 100 x 10⁵ )g  -  ( 4.701 x 10⁵) g

= (388 x 10⁵ )g -  ( 4.701 x 10⁵) g

= (388  - 4.701 ) x (10⁵ )g

= 383.299 x 10⁵ g

In standard form, the simplified expression can be expressed as;

= (3.83299 x 100 x 10⁵) g

= 3.83299 x 10⁷ g

= 3.833 x 10⁷ g

Therefore, the simplified expression is  3.833 x 10⁷ g

20. How much charge will flow through a 2002 galvanometer
connected to a 40092 circular coil of 1000 turns on a wooden
stick 2 cm in diameter? If a magnetic field B=0.011 T parallel to
the axis of the stick is decreased suddenly to zero?

Answers

Answer:

5.76 μC

Explanation:

The induce emf, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = NAΔB where N = number of turns of coil = 1000, A = cross-sectional area of coil = πd²/4 where d = diameter of coil = 2 cm = 2 × 10⁻² m and ΔB = change in magnetic field strength = B' - B where B' = final magnetic field = 0 T and B = initial magnetic field strength = 0.011 T. So, ΔB = 0 T - 0.011 T = -0.011 T

So, ε = -ΔΦ/Δt

ε = -NAΔB/Δt

ε = -NAΔB/Δt

Also ε = iR where i = current and R = combined resistance of circular coil and galvanometer = 200 Ω + 400 Ω = 600 Ω (since they are in series)

So, iR = -NAΔB/Δt

iΔt = -NAΔB/R

Δq = -NAΔB/R where Δq = charge = iΔt

substituting the values of the variables into the equation, we have

Δq = -1000 × π(2 × 10⁻² m)²/4 × -0.011 T/600 Ω

Δq = -1000 × 4π × 10⁻⁴ m²/4 × -0.011 T/600 Ω

Δq = 0.011π × 10⁻¹ m²T/600 Ω

Δq = 0.03456 × 10⁻¹ m²T/600 Ω

Δq = 5.76 × 10⁻⁶ C

Δq = 5.76 μC

A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes

Answers

Answer:

T1 = 499.9N, T2 = 865.8N, T3 = 1000N

Explanation:

To find the tensions we need to find the vertical and horizontal components of T1 and T2

T1x = T1 cos60⁰, T1y = T1 sin60⁰

Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰

For the forces to be in equilibrium,

the sum of vertical forces must be zero and the sum of horizontal forces must also be zero

Sum of Fx = 0

That is, T1x - T2x=0

NB: T2x is being subtracted because T1x and T2x are in opposite directions

T1 cos60⁰ - T2 cos30⁰ = 0

0.866T1 - 0.5T2 = 0 ............ (1)

Sum of Fy = 0

T1y + T2y - 1000 = 0

T1 sin60⁰ + T2 sin30⁰ - 1000 = 0

NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.

0.5T1 - 0.866T2 - 1000 = 0 ........(2)

From (1)

make T1 the subject

T1 = 0.5T2/0.866

Substitute T1 into (2)

0.5 (0.5T2/0.866) - 0.866T2 = 1000

(0.25/0.866)T2 - 0.866T2 = 1000

0.289T2 - 0.866T2 = 1000

1.155T2 = 1000

T2 = 865.8N

Then T1 = 0.5 x 865.8 / 0.866

T1 = 499.9N

T3 = 1000N

NB: The weight of the bag is the Tension above the rope, which is T3

explain why sound wave travel faster in liquid than gas​

Answers

Answer:

Because gas contains free molecules but not liquid.

Please mark as brainliast

why is it wrong to leave our light on​

Answers

Answer:

you will get huge electricity bills ............

why does self inductance acts as electrical inertia?​

Answers

Answer:

self-indulgence of coil is the property by virtue of wich is tends to maintain magnatic flux link with it and opposed any any change in the flux inducing current in it

A block of mass M is connected by a string and pulley to a hanging mass m.The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. Find the acceleration of the system and tensions on the string.

Answers

The free body diagram for the block of mass M consists of four forces:

• the block's weight, Mg, pointing downward

• the normal force of the table pushing upward on the block, also with magnitude Mg

• kinetic friction with magnitude µMg = 0.2 Mg, pointing to the left

• tension of magnitude T pulling the block to the right

For the block of mass m, there are only two forces:

• its weight, mg, pulling downward

• tension T pulling upward

The m-block will pull the M-block toward the edge of the table, so we take the right direction to be positive for the M-block, and downward to be positive for the m-block.

Newton's second law gives us

T - 0.2Mg = Ma

mg - T = ma

where a is the acceleration of either block/the system. Adding these equations together eliminates T and we can solve for a :

mg - 0.2 Mg = (m + M) a

a = (m - 0.2M) / (m + M) g

a = 1.96 m/s²

Then the tension in the string is

T = m (g - a)

T = 78.4 N

If 5.4 J of work is needed to stretch a spring from 15 cm to 21 cm and another 9 J is needed to stretch it from 21 cm to 27 cm, what is the natural length (in cm) of the spring

Answers

Answer:

the natural length of the spring is 9 cm

Explanation:

let the natural length of the spring = L

For each of the work done, we set up an integral equation;

[tex]5.4 = \int\limits^{21-l}_{15-l} {kx} \, dx \\\\5.4 = [\frac{1}{2}kx^2 ]^{21-l}_{15-l}\\\\5.4 = \frac{k}{2} [(21-l)^2 - (15-l)^2]\\\\k = \frac{2(5.4)}{(21-l)^2 - (15-l)^2} \ \ \ -----(1)[/tex]

The second equation of work done is set up as follows;

[tex]9 = \int\limits^{27-l}_{21-l} {kx} \, dx \\\\9 = [\frac{1}{2}kx^2 ]^{27-l}_{21-l}\\\\9 = \frac{k}{2} [(27-l)^2 - (21-l)^2] \\\\k = \frac{2(9)}{(27-l)^2 - (21-l)^2} \ \ \ -----(2)[/tex]

solve equation (1) and equation (2) together;

[tex]\frac{2(9)}{(27-l)^2 - (21-l)^2} = \frac{2(5.4)}{(21-l)^2 - (15-l)^2}\\\\\frac{2(9)}{2(5.4)} = \frac{(27-l)^2 - (21-l)^2}{(21-l)^2 - (15-l)^2}\\\\\frac{9}{5.4} = \frac{(729 - 54l+ l^2) - (441-42l+ l^2)}{(441-42l+ l^2) - (225 -30l+ l^2)} \\\\\frac{9}{5.4 } = \frac{288-12l}{216-12l} \\\\\frac{9}{5.4 } =\frac{12}{12} (\frac{24-l}{18 -l})\\\\\frac{9}{5.4 } = \frac{24-l}{18 -l}\\\\9(18-l) = 5.4(24-l)\\\\162-9l = 129.6-5.4l\\\\162-129.6 = 9l - 5.4 l\\\\32.4 = 3.6 l\\\\l = \frac{32.4}{3.6} \\\\[/tex]

[tex]l = 9 \ cm[/tex]

Therefore, the natural length of the spring is 9 cm

How much work is required to stretch an ideal spring of spring constant (force constant) 40 N/m from x

Answers

Answer:

The work done will be "0.45 J".

Explanation:

Given:

K = 40 N/m

x₁ = 0.20 m

x₂ = 0.25 m

Now,

The required work done will be:

= [tex]\frac{1}{2}k[x_2^2-x_1][/tex]

By putting the values, we get

= [tex]\frac{40}{2}[(0.25)^2-(0.20)^2][/tex]

= [tex]20\times 0.0225[/tex]

= [tex]0.45 \ J[/tex]

Pure water is an example of alan
A. Insulator
B. Metalloid
C. Conductor
D. Nonmetal

Answers

A. Insulator is the right answer

Answer: I think its A or C I'm not sure though sorry.

4. Consider a 1 kg block is on a 45° slope of ice. It is connected to a 0.4 kg block by a cable
and pulley. Does the 1 kg block move or down the slope? What is the net force on it and
its acceleration? (8 pts)

Answers

If an icy surface means no friction, then Newton's second law tells us the net forces on either block are

• m = 1 kg:

F (parallel) = mg sin(45°) - T = ma … … … [1]

F (perpendicular) = n - mg cos(45°) = 0

Notice that we're taking down-the-slope to be positive direction parallel to the surface.

• m = 0.4 kg:

F (vertical) = T - mg = ma … … … [2]

Adding equations [1] and [2] eliminates T, so that

((1 kg) g sin(45°) - T ) + (T - (0.4 kg) g) = (1 kg + 0.4 kg) a

(1 kg) g sin(45°) - (0.4 kg) g = (1.4 kg) a

==>   a ≈ 2.15 m/s²

The fact that a is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is a ≈ 2.15 m/s², which means the net force on the block would be ∑ F = ma ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.

When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.

Answers

Answer:

angular acceleration.

Explanation:

Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.

Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.

A nerve impulse travels along a myelinated neuron at 90.1 m/s.
What is this speed in mi/h?

Answers

Answer:

201.5537 mph

Explanation:

Given the following data;

Speed = 90.1 m/s

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

Speed = distance/time

To convert this value into miles per hour;

Conversion;

1 meter = 0.000621 mile

90.1 meters = 90.1 * 0.000621 = 0.05595 miles

1 metre per second = 2.237 miles per hour

90.1 meters per seconds = 90.1 * 2.237 = 201.5537 miles per hour

90.1 m/s = 201.5537 mph

A horizontal, mass spring system undergoes simple harmonic motion. which of the following statements is correct reguarding the mass in the system when it is located at its maximum distance from the equilibrium position?
a. The acceleration of the mass is zero.
b. The potential energy of the spring attached to the mass is at a minimum.
c. The total mechanical energy of the mass is zero.
d. The kinetic energy of the mass is at a maximum.
e. The speed of the mass is zero.

Answers

Answer:

Option (e)

Explanation:

A body executing SHM moves to and fro or back and forth  about its mean position.

When the particle is at mean position, its velocity is maximum and when it is at extreme position its velocity is zero.

So, when it is at maximum distance:

a.

The acceleration is maximum.

b.

The potential energy is maximum.

c.

The total mechanical energy is non zero.

d.

The kinetic energy is zero.

e. The speed is zero. Correct

A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of the point charge and the total charge on shell were, respectively:

Answers

Complete question is;

A point charge is positioned in the center of a hollow metallic shell of radius R. During four experiments the value of point charge and charge of the shell were respectively:

+5q; 0

-6q; +2q

+2q; -3q

-4q; +12q

Rank the results of experiments according to the charge on the inner surface of the shell, most positive first:

a. 2, 3, 1, 4

b. 1, 2, 3, 4

c. 2, 4, 3, 1

d. 1, 3, 4, 2

Answer:

c. 2, 4, 3, 1

Explanation:

In this question, we can say that;

q_in = q_b

Where;

q_in is the charge on the inner surface of the shell

q_b is the point charge on the shell.

Thus q_in = -q_b was written because, as the shell is conducting, it means that the electric field would have a value of zero and thus the radius inside will be zero.

Thus;

- For +5q; 0:

q_in = -(+5q)

q_in = -5q

- For -6q; +2q :

q_in = - (-6q)

q_in = +6q

- For +2q; -3q :

q_in = -(+2q)

q_in = -2q

- For -4q; +12q:

q_in = -(-4q)

q_in = +4q

Ranking the most positive to the least positive ones, we have;

+6q, +4q, -2q, -5q

This corresponds to options;

2, 4, 3, 1

In a
DC source, which has more cuwent?
(i)R L Circuit
(ii)RC Circuit (series)
(iii)LC Cirenit (series)
(iv)RLC Circuit (series)

Answers

Answer:

Answer is LC Cirenit (seres)

A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18 m/s in the x-direction, 24 m/s in the y-direction, and 72 m/s in the z-direction. The segment has length 0.50 m and is parallel to the z-axis as it moves.

Required:
a. Find the motional emf induced between the ends of the segment.
b. What would the motional emf be if the wire segment was parallel to the y-axis?

Answers

Answer:

Explanation:

From the information given:

The motional emf can be computed by using the formula:

[tex]E = L^{\to}*(V^\to*\beta^{\to})[/tex]

[tex]E = L^{\to}*((x+y+z)*\beta^{\to})[/tex]

[tex]E = 0.50*((18\hat i+24 \hat j +72 \hat k )*0.0800)[/tex]

[tex]E = 0.50*((18*0.800)\hat k +0j+(72*0.080) \hat -i ))[/tex]

[tex]E = 0.50*((18*0.800)[/tex]

E = 0.72 volts

According to the question, suppose the wire segment was parallel, there will no be any emf induced since the magnetic field is present along the y-axis.

As such, for any motional emf should be induced, the magnetic field, length, and velocity are required to be perpendicular to one another .

Then the motional emf will be:

[tex]E = 0.50 \hat j *((18*0.800)\hat k -(72*0.080) \hat i ))[/tex]

E = 0 (zero)

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