If you were unfortunate enough to be 5.5 mm away from such a lightning bolt, how large a magnetic field would you experience

Answers

Answer 1

Answer:

The question is incomplete. Here is the complete question.

Lightning bolts can carry currents up to approximately 20kA. We can model such a current as the equivalent of a very long, straight wire.

(a) If you were unfortunate enough to be 5.5m away from such a lightning bolt, how large a magnetic field would you experience?

(b) How does this field compare to one you would experience by being 5.5cm from a long, straight household current of 5A?

Answer: (a) B = 7.27 x 10⁻⁴ T

(b) Approximately 40 times higher than a household one.

Explanation: Using Biot-Savart Law, the magnetic field in a straight, long wire is given by

$B=(\mu_(0)I)/(2.\pi.r)$

where:

$\mu_(0)$ (permeability of free space) = $4.\pi.10^(-7)$ T.m/A

(a) If lightning bolt is compared to a long and straight wire, then magnetic field is

$B=(4.\pi.10^(-7).10.10^(3))/(2.\pi.5.5)$

B = 7.27 x 10⁻⁴ T

The magnitude of magnetic field in a lightning bolt is 7.27 x 10⁻⁴ T

(b) Magnetic field in a household wire will be

$B=(4.\pi.10^(-7).5)/(2.\pi.5.5.10^(-2))$

B = 1.82 x 10⁻⁵ T

Comparing fields:

$(7.27.10^(-4))/(1.82.10^(-5))$ ≈ 40

The filed for a lightning bolt is approximately 40 times higher than for a household wire.

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The magnetic flux that passes through one turn of a 11-turn coil of wire changes to 5.60 from 9.69 Wb in a time of 0.0657 s. The average induced current in the coil is 297 A. What is the resistance of the wire?

Answers

Answer:

2.31 Ω

Explanation:

According to the Faraday's law of electromagnetic induction,

Induced emf = - N (dΦ/dt)

Emf = -N (ΔΦ/t)

where N = number of turns = 11

Φ = magnetic flux

ΔΦ = change in magnetic flux = 9.69 - 5.60 = 4.09 Wb

t = time taken for the change = 0.0657 s

Emf = 11(4.09/0.0657)

Emf = - 684.78 V (the minus sign indicates that the direction of the induced emf is opposite to the direction of change of magnetic flux)

From Ohm's law,

Emf = IR

R = (Emf)/I

I = current = 297 A

R = (684.78)/297

R = 2.31 Ω

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Explanation:

Below is an attachment containing the solution.

A solid 0.6950 kg ball rolls without slipping down a track toward a vertical loop of radius ????=0.8950 m . What minimum translational speed ????min must the ball have when it is a height H=1.377 m above the bottom of the loop in order to complete the loop without falling off the track? Assume that the radius of the ball itself is much smaller than the loop radius ???? . Use ????=9.810 m/s2 for the acceleration due to gravity.

Answers

Answer:

The minimum transnational speed is 4.10 m/s.

Explanation:

Given that,

Mass of solid ball = 0.6950 kg

Radius = 0.8950 m

Height = 1.377 m

We need to calculate the minimum velocity of the ball at bottom of the loop to complete the track

Using formula velocity at lower point

$v_(min)=√(5gR)$

Put the value into the formula

$v_(min)=√(5*9.8*0.8950)$

$v_(min)=6.62\ m/s$

We need to calculate the velocity

Using conservation of energy

P.E at height +K.E at height = K.E at the bottom

$mgH+(1)/(2)mv^2=(1)/(2)m(√(5gR))^2$

$v^2=(√(5gR))^2-2gH$

$v^2=(6.62)^2-2*9.8*1.377$

$v^2=16.8352$

$v=√(16.8352)$

$v=4.10\ m/s$

Hence, The minimum transnational speed is 4.10 m/s.

Final answer:

The minimum translational speed the solid ball must have when it is at a height H=1.377 m above the bottom of the loop to successfully complete the loop without falling off the track is approximately 7.672 m/s. This was derived using principles of energy conservation.

Explanation:

The minimum translational speed must be sufficient enough to maintain contact with the track even at the highest point of the loop. Using the principle of energy conservation, the total energy at the height H, assuming potential energy to be zero here, should be equal to the total energy at the highest point of the loop. Here, the total energy at height H will consist of both kinetic and potential energy while at the top of the loop it consists of potential energy only. Setting these equations equal to each other: 0.5 * m * v² + m * g * H = m * g * 2R Solving the above equation for v:v = √2g (2R-H). Substituting known values henceforth gives us √2*9.81*(2*0.895-1.377) = 7.672 m/s. Hence, the ball must have a minimum translational speed of approximately 7.672 m/s at height H to complete the loop without falling.

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Two long ideal solenoids (with radii 20 mm and 30 mm, respectively) have the same number of turns of wire per unit length. The solenoid is mounted inside the larger, along a common axis. The magnetic field with in the inner solenoid is zero. The current in the inner solenoid must be: a. two-thirds the current in the outer solenoid
b. one-third the current in the outer solenoid
c. twice the current in the outer solenoid
d. half of the current in the outer solenoid
e. the same as the current in the outer solenoid

Answers

Answer: The current in the inner solenoid is the same as the current in the outer solenoid.

The correct option is e

Explanation: Please see the attachment below

A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. Frope W =

Answers

We have that for the Question it can be said that the magnitude of the force exerted by the horizontal rope on her arms and the ratio of the Force to the weight is

F=1150.561N

F/W=1.8325

From the question we are told

A 64.0-kg ice skater is moving at 4.04 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.890 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. kN (b) Compare this force with her weight. F-rope W =

Generally the equation for the force applied is mathematically given as

$F=(( mv^2))/(R)\n\nTherefore\n\nF=(( mv^2))/(R)\n\nF=(( (64)(4.0)^2))/(0.890)\n\n$

F=1150.561N

Generally the equation for the Weight is mathematically given as

W=mg

Therefore

W=64*9.81

W=627.84N

Therefore

The Force to weight ratio is

$F/W=1150.561N/627.84N$

F/W=1.8325

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Final answer:

The force exerted by the rope on the skater's arms as she moves in a circular path is 1.167 kN. This force is about 1.860 times her weight, which is 627.2 N.

Explanation:

The skater is experiencing centripetal force exerted by the rope, which causes her to move in a circular path. The magnitude F of this force can be calculated using the formula F = mv²/r, where m is the skater's mass (64.0 kg), v is her velocity (4.04 m/s), and r is the radius of her circular path (0.890 m).

By substituting the given numbers into this formula, we get: F = (64.0 kg)(4.04 m/s)² / 0.890 m = 1166.67 N. In kilonewtons, this force is 1.167 kN.

To compare this force with her weight, we can calculate the weight (W) using the formula W = mg, where g is the acceleration due to gravity (around 9.8 m/s²). Substituting the given mass into this formula gives us: W = (64.0 kg)(9.8 m/s²) = 627.2 N.

Comparing these two forces shows that the force exerted by the rope on her arms is about 1.860 times her weight.

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You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

$a \: = (v \: - u)/(t) \n \n a = (13.41 - 31.29)/(5) \n \n a \: = ( - 17.88)/(5) \n \n$

a = –3.576 m/s²

Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

$v \: = u \: + at \n 0 \: = 31.29 \: + \: ( - 3.576 * t) \n 0 \: = 31.29 \: - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \: = - 3.576 * t \n - 31.29 \: = - 3.576 * t \n divide \: both \: side \: by \: - 3.576 \n t \: = (- 31.29)/(- 3.576) \n$

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

${v}^(2) = {u}^(2) + 2as \n {0}^(2) = {31.29}^(2) + (2 * - 3.576 * s) \n 0 = 979.0641 - 7.152 s \n collect \: like \: terms \n 0 - 979.0641 = - 7.152 s \n - 979.0641 = - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s = (- 979.0641)/(- 7.152) \n \n$

s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

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Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=(v-u)/(t)$

$a=(13.41-31.29)/(5)=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576* t$

$t=(31.29)/(3.576)=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2* (-3.576)\cdot s$

s=136.89 m

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of

Answers

Complete Question

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = (1)/(2)mv^2$

Making v the subject

$v = \sqrt{[(2 \Delta V * q )/(m)] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{(2 * 5.9 *10^3 * 1.60218*10^(-19) )/(9.10939 *10^(-31]) )$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $(mv^2)/(r)$ ] and this is mathematically represented as