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The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?

Answers

Answer 1

Answer:

Answer:

The charge that passes through the starter motor is $\Delta Q=158.1 C$ .

Explanation:

Known Data

Avogadro's Number $N_(A)=6.02x10^(23)$
Current, $I=170A=170(C)/(s)$
Charge in an electron, $q=1.60x10^(-19)C$
Time, $\Delta t=0.930s$
Diameter, $d=4.60mm=0.0046m$
Transversal Area, $A=((d)/(2))^(2) \pi=((0.0046m)/(2))^(2) \pi=1.66x10^(-5) m^(2)$
Volume, $V=Length*A=(1.2m)(1.66x10^(-5) m^(2))=1.99x10^(-5) m^(3)$

First Step: Find the number of the electrons per unit of volume in the wire

We use the formula $n=(N_(A))/(V)= (6.02x10^(23) electrons)/(1.99x10^(-5) m^(3)) =3.02x10^(28)el/ m^(3)$ .

Second Step: Find the drag velocity

We can use the following formula $v_(d)=(I)/(nqA)=(170C/s)/((3.02x10^(28)m^(-3))(1.60x10^(-19)C)(1.66x10^(-5) m^(2))) =2.11x10^(-3) m/s$

Finally, we use the formula $\Delta Q=(nAv_(d)\Delta t)q=(3.02x10^(28) m^(-3))(1.66x10^(-5) m^(2))(2.11x10^(-3) m/s)(0.930s)(1.60x10^(-19)C)=158.1 C$ .

Related Questions

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Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?
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If i throw up an object up at 31 m/s, how long will it take to get to its highest point
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 acceleration due to gravity on Earth is 9.80 m/s2. If the mass of a honeybee is 0.000100 kilograms, what is the weight of this insect?

Answers

Answer:

0.00098 N

Explanation:

The weight of an object is given by:

$W=mg$

where

m is the mass of the object

g is the gravitational acceleration on the planet

In this problem, we have:

$m=0.0001 kg$ is the mass of the honeybee

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting into the equation, we find:

$W=mg=(0.0001 kg)(9.8 m/s^2)=0.00098 N$

weight = mg
here m = 0.000100 g = 9.80
hence weight = 0.00980 kgm/s2

A turntable with a rotational inertia of 0.0120 kg∙m2 rotates freely at 2.00 rad/s. A circular disk of mass 200 g and radius 30.0 cm, and initially not rotating, slips down a spindle and lands on the turntable. (a) Find the new angular velocity. (b) What is the change in kinetic energy?

Answers

To solve this problem it is necessary to apply the related concepts to the moment of inertia in a disk, the conservation of angular momentum and the kinematic energy equations for rotational movement.

PART A) By definition we know that the moment of inertia of a disk is given by the equation

$I = (1)/(2) MR^2$

Where

M = Mass of the disk

R = Radius

Replacing with our values we have

$I = (1)/(2) (0.2)(0.3)^2$

$I = 9*10^(-3)kg\cdot m^2$

The initial angular momentum then will be given as

$I = I_1 \omega_1$

$I = 0.012*2$

$I = 0.024kg\cdot m^2/s$

Therefore the total moment of inertia of the table and the disc will be

$I_2 = 9*10^(-3)+0.012$

$I_2 = 0.021kg\cdot m^2$

The angular velocity at the end point will be given through the conservation of the angular momentum for which it is understood that the proportion of inertia and angular velocity must be preserved. So

$I_1 \omega_1 = I_2\omega_2$

$(0.012)(2)=(1.08*10^(-4))\omega_2$

$\omega_2 = (0.012*2)/(0.021)$

$\omega_2 = 1.15rad/s$

Therefore the new angular velocity is 1.15rad/s

PART B) Through the conservation of rotational kinetic energy we can identify that its total change is subject to

$\Delta KE = (1)/(2)I_1\omega_1^2-(1)/(2)I_2\omega^2$

$\Delta KE = (1)/(2)(I_1\omega_1^2-I_2\omega^2)$

$\Delta KE = (1)/(2)(0.024*2^2-0.021*1.15^2)$

$\Delta KE = 0.034J$

Therefore the change in kinetic energy is 0.034J

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the cyclist begins to uniformly apply the brakes. the bicycle stops in 5.0 s. how far did the bicycle travel during the 5.0 seconds of braking?

Answers

Distance traveled by the bicycle during the 5 seconds of braking is 22m

Explanation:

initial angular velocity= 2 rev/s

final angular velocity= 0 rev/s

Angular displacement Ф= $((wi+wf)/(2) )$ t

Ф= $((0+2)/(2) )5=5$ rev

so the distance travelled= 5(2πr)

distance=5(2π*0.7)

distance=22m

The bicycle traveled about 22 m during the 5.0 seconds of braking

$\texttt{ }$

Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

$\texttt{ }$

Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

$\texttt{ }$

Given:

radius of wheel = R = 0.70 m

initial angular speed = ω = 2.0 rev/s = 4π rad/s

final angular speed = ωo = 0 rad/s

time taken = t = 5.0 s

Asked:

distance covered = d = ?

Solution:

$d = \theta R$

$d = (\omega + \omega_o)(1)/(2)t R$

$d = ( 4 \pi + 0 ) (1)/(2)(5.0)( 0.70 )$

$d = 4\pi (1.75)$

$d = 7\pi \texttt{ m}$

$d \approx 22 \texttt{ m}$

$\texttt{ }$

Learn more

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

The Richter scale is used to determine how strong the earthquake is (magnitude) of the earthquake. It touches the ground and feels the earth shaking. With the momentum of the earth shaking the device the needle on the device moves creating a wave looking line. According to the theory of plate tectonics, what happens at transform, divergent and convergent boundaries? On which of these boundary types would a volcano most likely take place, and why?

Answers

convergent, because it is where the tectonic plates shove themselves together usually resulting in a mountain or volcano

Final answer:

In plate tectonics theory, transform boundaries occur when two plates slide past each other, divergent boundaries occur when two plates move away from each other, and convergent boundaries occur when two plates collide. A volcano is most likely to occur at a divergent boundary because the plates move away from each other, allowing magma from the mantle to reach the surface and create new crust.

Explanation:

In plate tectonics theory, Transform boundaries occur when two plates slide past each other horizontally, creating earthquakes. Divergent boundaries occur when two plates move away from each other, creating volcanic activity.

Convergent boundaries occur when two plates collide, and depending on the type of plates involved, can result in volcanic activity as one plate is forced beneath the other.

A volcano is most likely to occur at a divergent boundary because the plates move away from each other, allowing magma from the mantle to reach the surface and create new crust.

The Richter scale indicates the magnitude of an earthquake. The figure drawn by the needle during shaking is an outcome of earthquake's energy. This energy is what results in seismic waves travelling through various layers of earth causing shaking on the surface.

Learn more about Plate tectonics and volcanoes here:

brainly.com/question/33604065

#SPJ12

If you travel 2 km north, then travel 5 km south, what is your displacement?

Answers

Answer:

➢This is a vector addition problem which requires magnitude and direction as the answer. First is to resolve the southbound vector and the northbound vector. Since they are opposite in directions their vector sum is their algebraic sum. 3 km north + 5 km south = 2 km south.

We then add 2 km west and 2 km south using Pythagorean theorem since west and south form a right angle. (2 km)^2 west + (2 km)^2 south gives (4 + 4) km^2 southwest = 8 (km)^2 45 degrees south of west

Extracting the square root of 8 gives us about 2.83 km 45 degrees south of west.

Explanation:

I hope it will help you...

The radius of Earth is 6370 km in the Earth reference frame. The cosmic ray is moving at 0.880Co relative to Earth.a. In the reference frame of a cosmic ray how wide does Earth seem along the flight direction?
b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?
Express your answer with the appropriate units.

Answers

Answer:

6052114.67492 m

$12.742* 10^(6)\ m$

Explanation:

v = Velocity of cosmic ray = 0.88c

c = Speed of light = $3* 10^8\ m/s$

d = Width of Earth = Diameter of Earth = $12.742* 10^(6)\ m$

When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original

This happens due to length contraction

Length contraction is given by

$d_e=d\sqrt{1-(v^2)/(c^2)}\n\Rightarrow d_e=12.742* 10^(6)\sqrt{1-(0.88^2c^2)/(c^2)}\n\Rightarrow d_e=6052114.67492\ m$

The Earth's width is 6052114.67492 m

Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.

Hence, the width is $12.742* 10^(6)\ m$

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Hotel rooms in Smalltown go for $100, and 1,000 rooms are rented on atypical day.a. To raise revenue, the mayor decides to charge hotels a tax of $10 perrented room. After the tax is imposed, the going rate for hotel roomsrises to $108, and the number of rooms rented falls to 900. Calculate theamount of revenue this tax raises for Smalltown and the deadweightloss of the tax. (Hint: The area of a triangle is l/2Xbase X height.)h. The mayor now doubles the tax to $20. The price rises to $116, andthe number of rooms rented falls to 800. Calculate tax revenue anddeadweight loss with this larger tax. Are they double, more thandouble, or less than double? Explain.

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