What is the maximum centripetal acceleration experienced by a person standing still on the surface of the Earth? Where must they be located?

Answers

Answer 1

Answer:

The person must be located in the Equator Line. The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

Explanation:

Physically speaking, the centripetal acceleration ([tex]a_{r}[/tex]), measured in meters per square second, experienced by a person is defined by the following expression:

[tex]a_{r} = \omega^{2}\cdot r[/tex] (1)

Where:

[tex]\omega[/tex] - Angular speed of the Earth, measured in radians per second.

[tex]r[/tex] - Distance perpendicular to the rotation axis, measured in meters.

Since rotation axis passes through poles and distance described above is directly proportional to centripetal acceleration. The person must be located in the Equator Line, which is equivalent to the radius of the planet.

In addition, the angular speed of the Earth can be calculated in terms of its period ([tex]T[/tex]), measured in seconds:

[tex]\omega = \frac{2\pi}{T}[/tex] (2)

If we know that [tex]r = 6.371\times 10^{6}\,m[/tex] and [tex]T = 86400\,s[/tex], then the maximum centripetal acceleration experienced by a person is:

[tex]a_{r} = \left(\frac{2\pi}{86400\,s} \right)^{2}\cdot (6.371\times 10^{6}\,m)[/tex]

[tex]a_{r} = 0.0337\,\frac{m}{s^{2}}[/tex]

The maximum centripetal acceleration experienced by a person is 0.0337 meters per square second.

Answer 2

The person standing still on the surface of the earth must be located in the equator line

Recall: the the centripetal acceleration at the Equator is about 0.03 m/s2.

This then means that the maximum centripetal acceleration of a person standing in the equator line is 0.03 m/s2

What is meant by maximum centripetal acceleration?

The maximum centripetal acceleration as the name implies is the maximum speed of a body or object in a circular path

Learn more about centripetal acceleration:

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Related Questions

A 0.25 kg beach ball rolling at a speed of 7 m/s collides with a heavy exercise ball at rest. The beach ball bounces straight back with a speed of 4 m/s. That is the change in momentum of the beach ball? What is the impulse exerted on the beach ball? What is the impulse exerted on the exercise ball?

Answers

The impulse exerted on the beach ball is 2.75 kgm/s.

The impulse exerted on the exercise ball is - 2.75 kgm/s.

What is impulse?

This is the force applied to an object that acts over a period of time.

The impulse exerted on the beach ball is the change in the momentum of the ball and it is calculated as follows;

J = ΔP

J = m(v - u)

J = 0.25(7 - (-4))

J = 0.25(7 + 4)

J = 2.75 kgm/s

The impulse exerted on the exercise ball is equal in magnitude but opposite in direction to the beach ball.

Thus, the impulse exerted on the exercise ball is - 2.75 kgm/s.

Learn more about impulse here: https://brainly.com/question/25700778

A baseball is thrown horizontally from a cliff at 30 m/s and lands 7 seconds after the baseball was thrown. Calculate the horizontal AND vertical distance.

Answers

Answer:

The horizontal and vertical distances are x = 210 m and y = -240.35 m, respectively.

Explanation:

Using the equation of the displacement in the x-direction, we have:

(let's recall we have a constant velocity in this direction)

[tex]x=v_{ix}t[/tex]

Where:

v(ix) is the initil velocity in the x direction (v(ix) = 30 m/s)t is the time (t = 7 s)

[tex]x=30(7)[/tex]    

[tex]x=210\: m[/tex]    

Now, we need to use the equation of the displacement in the y-direction to find the vertical distance. Here we have an acceleration (g)

[tex]y=v_{iy}t-\frac{1}{2}gt^2[/tex]

Where:

v(iy) is the initial velocity at the y-direction. In this case, it will be 0t is the timeg is the acceleration of gravity (g=9.81 m/s²)

Then, the vertical position at 7 s is:

[tex]y=-\frac{1}{2}(9.81)(7)^2[/tex]

[tex]y=-240.35\: m[/tex]

Therefore, the horizontal and vertical distances are x = 210 m and y = -240.35 m, respectively. The minus sign means the negative value in the y-direction.

I hope it helps you!

Allison and Heather are going to conduct an experiment to see whether salt affects the growth of plants. They assemble five groups of identical plants and give the plants in each group water with a different salt concentration. What is the outcome variable (dependent variable) for their experiment?
A. Salt concentration in plant tissue
B. Salt concentration in plant water
C. Amount of water absorbed by plants
D. Average mass of plants in each group

Answers

Answer:b

Explanation:

Guess

Stored energy due to vertical position is known as
Elastic Potential energy
Vibrational energy
Kinetic energy
O Gravitational Potential energy
1
2
3
4
5

Answers

Answer: gravitational potential energy

Explanation:

.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum acceleration for these elevators during normal operation, so you decide to measure it by using your bathroom scale. While the elevator is at rest on the ground floor,you get in, put down your scale, and stand on it. The scale reads 50 kg. You continue standing on the scale when the elevator goes up, carefully watching the reading. During the trip to the 10th floor, the greatest scale reading was

Answers

Answer:

5.51 m/s^2

Explanation:

Initial scale reading = 50 kg  

assume the greatest scale reading = 78.09 kg

Determine the maximum acceleration for these elevators

At rest the weight is = 50 kg

Weight ( F ) = mg = 50 * 9.81 = 490.5 N

At the 10th floor weight = 78.09 kg

Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N

F = change in weight

Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)

50 * a = 275.61

Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2

your "A" never changes, while your "Y" changes depending on strength of gravity. is it mass or weight? ​

Answers

Answer:

A - mass. B - Weight

Explanation:

This is because weight varies with the strength of gravity. Mass is just the amount of matter in an object

Help plsssssssssss I write it 100 time no one answer

Answers

Answer:

1.93×10²⁸ s

Explanation:

From the question given above, the following data were obtained:

Number of electron (e) = 2×10²⁴

Current (I) = 10 A

Time (t) =?

Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:

1 e = 96500 C

Therefore,

2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e

2×10²⁴ e = 1.93×10²⁹ C

Thus, 1.93×10²⁹ C of electricity is passing through the point.

Finally, we shall determine the time. This can be obtained as follow:

Current (I) = 10 A

Quantity of electricity = 1.93×10²⁹ C

Time (t) =?

Q = it

1.93×10²⁹ = 10 × t

Divide both side by 10

t = 1.93×10²⁹ / 10

t = 1.93×10²⁸ s

Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point

Given that Carbon-14 has a half-life of 5700 years, determine how long it would take for
this reduction to occur.

Answers

Answer:It will take about 3000 years

Explanation:

walking dancing and even some household chores are? ​

Answers

Answer:

Actions

Explanation:

if u mean what are these called then it's actions

Answer: Regular Physical Activity

Explanation:

Which is an example of kinetic energy?

A. The energy stored in
ethanol

B. A ball sitting at the top of a ramp

C. A compressed spring

D. A hockey puck sliding across ice

Answers

D . A hockey puck sliding across ice

D. A hockey puck sliding across ice

The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is [01] m. (a) What is the force (N) that the track must exert on the car? (positive is up) (b) What must be the force (N) that the car exerts on a 61 kg passenger?

Answers

This question is incomplete, the complete question is;

The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.

(a) What is the force (N) that the track must exert on the car? (positive is up)

(b) What must be the force (N) that the car exerts on a 61 kg passenger?

Answer:

a) the force (N) that the track must exert on the car is -6139.14 N

b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

Explanation:

Given the data in the question;

Let N represent the force that the track must exerted on the car

Net force on the car Fnet = Mg + N

so

M × a = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

we substitute

N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )

N = 3136 - ( 320 × 28.9848 )

N = 3136 - 9275.136

N = -6139.14 N

Therefore, the force (N) that the track must exert on the car is -6139.14 N

b) What must be the force (N) that the car exerts on a 61 kg passenger?

Let N represent the force that the car exerts on 61kg passengers

so

Net force of passengers Fnet = mg + N

Ma = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )

N = 597.8 - ( 61 × 28.9848)

N = 597.8 - 1768.0728

N = -1170.27 N

Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

The centripetal force of the track on the car moving in the circular path is [tex]1.465 \times 10^6 \ N[/tex].

The force (N) that the car exerts on a 61 kg passenger is 597.8 N.

Centripetal force of the track

The centripetal force of the track on the car moving in the circular path is calculated as follows;

[tex]F_c = \frac{mv^2}{r}\\\\ F_c = \frac{320 \times 21.4^2}{0.1} \\\\F_c = 1.465 \times 10^6 \ N[/tex]

Normal force of the passenger

The force (N) that the car exerts on a 61 kg passenger is equal to the force the passenger exerts on the car based on Newton's third law of motion.

F = mg

F = 61 x 9.8

F = 597.8 N

Learn more about centripetal force here: https://brainly.com/question/20905151

A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy is closer to the center. Who has greater angular displacement?
a) boy
b) girl
c) both have the same angular displacement

Answers

Answer:

c) both have the same angular displacement

Explanation:

In this scenario, girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy is closer to the center. Therefore, both have the same angular displacement.

Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.

Which describes the length of the power cord when Shanti gets off the train?

cannot be determined
less than 1.2 m
more than 1.2 m
equal to 1.2 m

Answers

Answer:

D. equal to 1.2

Explanation:

on edg

The length of the power cord will be equal to 1.2 m.

Describe about the length of power cord? The train is moving at a speed of 90 km /hr. Train was moving but the person in the train can be considered to be at rest. Shanti is the person travelling on the train. Her cord can be used only by her and the cord length of the phone will be 1.2 m.The length can be measured through the distance.The unit of length is meter.As we know the concept of motion and rest, there only the train in motion, shanti was at rest and shanti's power cord were also in the rest. Power cord length will be determined only at the time of manufacturing.If the power cord length to be change then the crimping process.So, the length will not change suddenly.

The length of the power cord when shanti gets off the train is equal to 1.2 m.

The Correct answer is Option D.

Learn more about motion and rest,

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Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be +(30±5)pC+(30±5)pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain). What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)?
(a) Because air is a good conductor, the positive charge on the bee’s surface flowed through the air from bee to plant.
(b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee.
(c) The plant became electrically polarized as the charged bee approached.
(d) Bees that had visited the plant earlier deposited a positive charge on the stem.

Answers

Answer:

a) True

Explanation:

There are several possible explanations for this positive charge

* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,

to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct

* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.

When reviewing the different statements we have

a) True, it agrees with the second explanation of the phenomenon

b) False. The earth is a deposit of negative charge

c) false. If this is the case the charge should be negative

d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.

Answer:

Explanation:

.

A student is conducting an experiment to compare the resistivity of two unknown materials by using two wires, each made of one of the materials and each connected in a circuit. The student measures the potential difference across and current in the wires. What must be the same to be able to compare the resistivities using just the potential difference and current measurements?

Answers

Answer:

is there a. b.  c  or d?

Explanation:

The First Law of Thermodynamics is the same as ______ with heat and work taken into consideration.

A. The First Law of Robotics
B. The Law of Conservation of Energy
C. Newton's First Law of Motion
D. The Law of Conservation of Momentum​

Answers

Answer:

the law of conservation of energy

Please help 25 points!

Three waves with frequencies of 1 Hertz (Hz), 3 Hz, and 9Hz travel at the same speed. Which of the following statements is correct?

A. The 1 Hz wave contains the most energy.

B. The crests of all three waves are of equal height.

C. The wavelength of the 9Hz wave is three times that of the 3 Hz wave.

D. The 1 Hz wave has the longest wavelength.

Answers

Answer:

B

Explanation:

The crest of all three waves are of equal height

3. Sodium-24 has a half-life of 15 hours. If a sample of sodium-24 has an
original activity of 800 Bq, what will
its activity be after:
i) 15 hours?
ii) 30 hours?
iii) 45 hours?
iv) 60 hours?

Answers

Answer:

See explanation

Explanation:

From the formula;

0.693/t1/2 = 2.303/t log (Ao/A)

t1/2 = half life of Sodium-24

Ao = initial activity of Sodium-24

A= activity of Sodium-24 at time = t

So,

0.693/15 = 2.303/15 log (800/A)

0.0462 = 0.1535  log (800/A)

0.0462/0.1535 =  log (800/A)

0.3 = log (800/A)

Antilog(0.3) =  (800/A)

1.995 =  (800/A)

A = 800/1.995

A = 401 Bq

ii) 0.693/15 = 2.303/30 log (800/A)

0.0462 = 0.0768 log (800/A)

0.0462/0.0768 =  log (800/A)

0.6 =  log (800/A)

Antilog (0.6) =  (800/A)

3.98 =  (800/A)

A = 800/3.98

A = 201 Bq

iii)

0.693/15 = 2.303/45 log (800/A)

0.0462 = 0.0512  log (800/A)

0.0462/0.0512  =  log (800/A)

0.9 = log (800/A)

Antilog (0.9) =  (800/A)

7.94 = (800/A)

A = 800/7.94

A= 100.8 Bq

iv)

0.693/15 = 2.303/60 log (800/A)

0.0462 = 0.038 log (800/A)

0.0462/0.038 = log (800/A)

1.216 = log (800/A)

Antilog(1.216) = (800/A)

16.44 = (800/A)

A = 800/16.44

A = 48.66 Bq

Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 0.55 cm above another. (The magnitude of this charge is consistent with what is typical of static electricity.)

Answers

Answer:

    q = 2 10⁻⁸ C

Explanation:

For this exercise we use the translational equilibrium equation

                    F_e -A =

                    F_e = W

the electric force is given by Coulomb's law

                    F_e = [tex]k \frac{q_1q_2}{r^2}[/tex]

in this case they indicate that the loads on the tapes are equal

                    F_e = k q² / r²

we substitute

                    k q² / r² = m g

                    q = [tex]\sqrt{ \frac{mg r^2}{k} }[/tex]

calculate  

                     q = [tex]\sqrt { \frac{ 12 \ 10^{-3} \ 9.8 (0.55 \ 10^{-2})^2 }{9 \ 10^9} }[/tex]    

                     q = [tex]\sqrt{ 3.9526 \ 10^{-16}[/tex]

                     q = 1,999 10⁻⁸ C

                     q = 2 10⁻⁸ C

TWO forces, one of 12N and another or 24N

act on body in such a way that they make an angle of 90° with each other. Find the resaltant of two forces.

Answers

Answer:

26.833 N

Explanation:

The computation of the resaltant of two forces is shown below:

Given that

Force A = 12N

Force B = 24N

Based on the above information  

Resultant R is

[tex]=\sqrt{A^2 + B^2 + 2AB \times cos \theta}\\\\=\sqrt{144 + 576 + 2\times 24\times 12\times cos90^{\circ}}\\\\=\sqrt{144+576+576\times 0}\\\\=\sqrt{720}[/tex]

=26.833 N

What would we need to do to make an electromagnet strong enough to move cars and trains

Answers

Answer:

The combined magnetic force of the magnetized wire coil and iron bar makes an electromagnet very strong. In fact, electromagnets are the strongest magnets made. An electromagnet is stronger if there are more turns in the coil of wire or there is more current flowing through it.

If the mass of the object doubles then the acceleration is when the force is kept the same

Answers

Answer:

Halved

Explanation:

F=ma

Let case 1 (original) be:

[tex]F_{1}=m_{1} a_{1} \\[/tex]

Case 2 (new) be:

[tex]F_{2}=m_{2} a_{2}[/tex]

Mass is double:

[tex]m_{2}= 2m_{1}[/tex]

Force kept the same:

[tex]F_{1} =F_{2}[/tex]

Combine the equation and gives:

[tex]\frac{F_{1} }{F_{2}} =\frac{m_{1} a_{1} }{m_{2}a_{2} }\\\frac{F_{1} }{F_{1}} =\frac{m_{1} a_{1} }{2m_{1}a_{2} }\\1=\frac{a_{1} }{2a_{2} }\\a_{2}=\frac{1}{2} a_{1}[/tex]

Acceleration is halved

A fan has four identical, symmetrically placed blades. The blades are rotating clockwise at twenty revolutions per second.


A) What is the smallest time interval between stroboscope flashes that will make the fan blades appear motionless?


B) What is the highest frequency (in flashes per second) at which a stroboscope will make the

fan blades appear to stand still? Show your calculation.


C) The same questions as (a) and (b), but someone has put a yellow dot on one blade, and now you want the yellow dot to appear to be standing still. Explain, and show your calculation.


D) Now the stroboscope is set for nineteen flashes per second, and the yellow dot appears to be slowly rotating. Which direction does it appear to rotate, clockwise or counterclockwise? Explain, and show your calculation.


E) The same as (d), but the stroboscope is set for twenty-one flashes per second. Explain, and show your calculation.

Answers

Answer:

A) t = 1.249 10⁻² s, B)  f = 80 Hz, C) f = 20 Hz,

D)  slowly advancing an angle of approximately    Δθ = 0.05 rad each flash

E) In each flash it seems to go backward an angle of Δθ = -0.05 rad

Explanation:

A) To make it appear that the blades are immobile, it implies that every time the light turns on, a blade should be in the same position, therefore, as we have 4 blades, they must rotate an angle of 2π/4,  

         θ = π / 2  

         θ = 1.57 rad  

taking the angle let's use the endowment kinematics relations  

          θ = w₀ t + ½ α t²  

in general the fans rotate at constant speed α= 0  

         θ = w₀ t  

         t = θ / w₀  

let's reduce the magnitudes to the SI system  

        w₀ = 20 rev / s (2π rad / 1rev) = 125.66 rad / s  

let's calculate  

        t = 1.57 / 125.66  

        t = 1.249 10⁻² s  

B) the fastest speed for the blades to rotate is when one blade of a complete turn , we use the relationship between the fecuance and the period  

        f = 1 / T  

        f = 1 / 1.25 10⁻²  

       f = 80 Hz

C) we have two possibilities:  

* a yellow dot is placed on each sheet  

In this case the angular velocity of the blade is the same at all points, therefore the results obtained should not change

* a yellow dot is placed on a single sheet.  

Here for the point to remain fixed the angle of rotation must be

       θ= 2π rad  

the time is  

       t = 2π / 125.66  

       t = 5 10⁻² s  

the maximum frequency is  

      f = 1/5 10⁻²  

      f = 20 Hz

D) The copy strobe rotates at f = 19 Hz, the time between each flash is  

      t = 1/19  

      t = 5.26 10⁻² s  

this time is higher, so the angle turned is large  

       θ = w t  

       θ = 125.66 5.26 10⁻²  

       θ = 6.61 rad  

the relationship between this angle and the angle of a circle is  

θ = 1,052

We can see that it is this time the blade rotates 1 complete turns, for this the position of the blade changes us, for the other 0.052 rad the blade rotates a little more than the circumference therefore it seems that it is slowly advancing an angle of approximately  

         Δθ = 0.05 rad each flash  

E) in this case changes the flash speed  

       t = 1/21  

       t = 4.76 10⁻² s  

the angle rotated is  

      θ = 125.66 4.76 10⁻²  

      θ = 5.984 rad  

      θ / 2π = 0.95  

in that case, the blade did not complete the turn, therefore in each flash it seems to go backward an angle

Δθ = -0.05 rad

if a current of 5A flows for 2minutes, find the quantity of electricity transfered ​

Answers

It’s =24c Because we divided two number and get it

Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 X10^-9 C and the other is given a charge of -18 X 10^-9 C. a. Find the electric force exerted on one sphere by the other. b. The sphere are connected by a conducting wire. After equilibrium has occurred, find the electric force between the two spheres.

Answers

Answer:

Explanation:

Force between two charged conducting sphere

= k x Q₁ x Q₂ / r² ,  k is a constant  Q₁ and Q₂ are charges and   r is distance between them .

= 9 x 10⁹ x 12 x 10⁻⁹ x 18 x 10⁻⁹ / .30²

= 21600 x 10⁻⁹

= 2.16 x 10⁻⁵ N .

b )

After the spheres are joined together , there is redistribution of charge and remaining charge will be equally shared by them .

Charge on each sphere = (12 - 18 ) x 10⁻⁹ / 2

= - 3 x 10⁻⁹ C .

Force = 9 x 10⁹ x 3 x 10⁻⁹ x 3 x 10⁻⁹ / .30²

= 900 x 10⁻⁹ N .

how many pennies can 4 folds of a paper hold?

Answers

A lot I don’t really under stand the question

If a car is moving to the left with constant velocity, one can conclude that :

a. there must be no forces exerted on the car.
b. the net force exerted on the car is directed to the left.
c. the net force exerted on the car is zero.
d. there is exactly one force exerted on the car.

Answers

Answer:

b. the net force exerted on the car is directed to the left.

Explanation:

Applying Newton's second law of motion, the car will move in the direction of the applied force. If the applied forces are on different directions, the car will move in the direction of the greater force (net force).

Therefore, if a car is moving to the left with constant velocity, one can conclude that the net force exerted on the car is directed to the left.

The correct option is "b. the net force exerted on the car is directed to the left"

Force = mass X acceleration. Acceleration due to gravity is 9.8 m/s/s regardless of height or mass. This means a single object will hit the ground with the same force regardless of the height it is dropped from. If this is true, I normally drop a pumpkin from different heights during this unit. why does dropping the pumpkin stay together when dropped from one meter but break apart when dropped from a height of 5 meters?

Answers

1 yaa I now homar and dog ririnebjeildnbbd

The current flow in the light bulb is 0.5A
a.Calculate the amount of electric charge that flow through the bulb in 2 hour
b.If one election carries a
charge 1.6 x 10-14 c Find the number of election through the bulb in 2 hour?

Answers

Answer:

Explanation:

Given that,

The current in the light bulb, I = 0.5 A

(a) We know that,

Electric current = charge/time

or

Q = It

Put t = 2 hours = 7200 s

So,

Q = 0.5 × 7200

Q = 3600 C

(b) Charge on one electron, [tex]Q=1.6\times 10^{-19}\ C[/tex]

Let there are n electrons flow through the bulb in 2 hours.

I = Q/t

Since, Q = ne

So,

I = ne/t

[tex]n=\dfrac{I\times t}{e}\\\\n=\dfrac{0.5\times 7200}{1.6\times 10^{-19}}\\\\n=2.25\times 10^{22}[/tex]

Hence, this is the required solution.

4 Two friction disks A and B are to be brought into contact withoutslipping when the angular velocity of disk A is 240 rpm counterclockwise. Disk A starts from rest at time t = 0 and is given a constantangular acceleration with a magnitude α. Disk B starts from rest attime t = 2 s and is given a constant clockwise angular acceleration,also with a magnitude α. Determine (a) the required angular acceleration magnitude α, (b) the time at which the contact occurs

Answers

This question is incomplete, the missing image is uploaded along this answer below;

Answer:

a) the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) the time at which the contact occur is 8 seconds

Explanation:

Given the data in the question;

first we convert the given angular velocity to rad/s

angular velocity = 240 rpm = ((240/60) × 2π ) = 8π rad/s

so

ωA = 8π rad/s

next we determine angular acceleration at point A; so

ωA = at

8π rad/s = at -------let this be equation

thus, angular acceleration of disk A is ωA and rotates in counter clockwise direction.

Next we determine the velocity of point C;

Vc = rA × ωA

where Vc is velocity at point C, rA is radius of A ( 150/1000)m,  { from the diagram }

so we substitute

Vc = 0.15m × 8π

Vc = 1.2π m/s

for angular velocity at point B;

Vc = rB × ωB

where rB is the radius of B ( 200/1000)m

we substitute

1.2π = 0.2 × ωB

ωB = 1.2π / 0.2

ωB = 6π rad/s

Thus, the wheel B rotates with an angular velocity of 6π rad/s in clock wise direction.

Now,

a) Determine the required angular acceleration magnitude α

we find the the angular acceleration of disk B after 2 seconds, using the expression;

ωB = at

where angular acceleration is a and t is time ( t - 2)

we substitute

ωB = at

6π = a( t - 2) -------- let this be equation 2

now, lets substract equation 1 form equation 2

(6π = a( t - 2)) - (8π = at)

(6π = at - 2a) - ( 8π = at)

-2π = 0 + -2a

2π = 2a

a = 2π/2

a = π rad/s² or 3.14 rad/s²

Therefore, the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) determine the time at which the contact occurs;

from equation 1

8π = at

we substitute in the value of a

8π = π × t

t = 8π / π

t = 8 seconds

Therefore, the time at which the contact occur is 8 seconds

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