I have a problem in this questions?

I Have A Problem In This Questions?

Answers

Answer 1

Answer:

8.46E+1

Explanation:

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

Charge 1 (q₁) = 39 C

Charge 2 (q₂) = –53 C

Force (F) of attraction = 26×10⁸ N

Electrical constant K) = 9×10⁹ Nm²/C²

Distance apart (r) =?

The distance between the two charges can be obtained as follow:

F = Kq₁q₂ / r²

26×10⁸ = 9×10⁹ × 39 × 53 / r²

26×10⁸ = 1.8603×10¹³ / r²

Cross multiply

26×10⁸ × r² = 1.8603×10¹³

Divide both side by 26×10⁸

r² = 1.8603×10¹³ / 26×10⁸

r² = 7155

Take the square root of both side

r = √7155

r = 84.6 m

r = 8.46E+1 m


Related Questions

The wire A and B are made of copper. Both w
wires are in
long but wire A is 1mm thick and wire B is 2mm thick, the
specific resistance is
a) more than A
b) more than B
c) same for both
d) cannot compared​

Answers

the specific resistance is more than B

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

What is the Radiation left over from the big bang called?

Answers

Answer:

The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe should be filled with radiation that is literally the remnant heat left over from the Big Bang, called the “cosmic microwave background", or CMB.

Explanation:

The Big Bang theory suggest  that the universe in early stage was at very hot place and which can be expanded, the gas within it cools. It is in an infinite universe and it has no edge.

What is big bang theory ?

The Big Bang theory is a cosmological model which explain the existence of the observable universe from the earliest periods to the large-scale evolution.

The model describes the mechanism behind the universe expansion from an initial state of high density and temperature, it is very important concept as a lot of research is going on in this field to find out exactly how the universe began billions of years ago.

The universe began to cool down in order to allow the formation of particles become atoms after its initial phase of expansion,  Primordial elements  such as Hydrogen, Helium, and Lithium are condensed through gravity are formed early stars and galaxies.

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In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are
?
A. the same
B. opposite
C. constantly changing.

Answers

Answer:

answer = opposite

they are not changing being in a place as being they-self

the same would not be true because its different then the Styrofoam and it could only be different and its make the most sense aswell.

Both ball and rod has opposite charges on their bodies because of presence of different charges on it.

What are the charges on ball and rod?

In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are opposite because Styrofoam ball is negatively charged due to the presence of electrons while on the other hand, the rod is positively charged because of lining of positive charges on the rod.

So we can conclude that both ball and rod has opposite charges on their bodies because of presence of different charges on it.

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Question 3 (5 points)
Yissel was going to be late to Mr. Scharff's science class. Just as the bell was about to ring. Vissel ran the last little bit of the hallway at 2.47
meters/second for 8 seconds to beat the bell. How far away was Yissel from Mr. Scharff's classroom when she started to run?

Answers

I am yissell what’s up by the way I was I was 10 meters of stairs from his room has to go down the stairs and wait to finish exit ticket

If you have a 0.125 kg lead piece at
20.0°C, how much heat must you
add to melt it? (Remember, you
must warm it to its melting point
first.)
Material
Lead
Melt Pt (°C)
327
L (1/kg)
2.32.104
Boil Pt (°C) Lv (1/kg)
1750 8.59.105
c (1/(kg*c)
128
(Unit = J)

Answers

Answer:

7,812 J

Explanation:

Using the relation:

Q = mcΔθ

Q = quantity of heat

C = specific heat capacity of lead

Δθ = temperature change (T2 - T1)

M = mass of substance

Q = mass * specific heat * Δθ

Q = 0.125kg * 128 * (327 – 20)

Q = 0.125 * 128 * 307

Q = 4912 J

For melting:

Q = mass * Hf

0.125 * (2.32 * 10^4)

= 2,900 J

Total = 4,912 J + 2,900 J = 7,812 J

A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.

Answers

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

A 450.0 kg roller coaster is traveling in a circle with radius 15.0m. Its speed at point A is 28.0m/s and its speed at point B is 14.0 m/s. At point A the cart is already moving with circular motion. a) Draw free bodydiagramsfor the cartatpointsAand B(two separate free body diagrams). b) Calculate the acceleration of the cartat pointsAandB(magnitude and direction). c) Calculate the magnitude of the normal force exerted by the trackson the cartat point A. d) Calculate the magnitude of the normal force exerted by the tracks on the cart at point B.

Answers

Answer:

b)  a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N

Explanation:

a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle

b) Let's start at point A

Let's use that the acceleration is centripetal

           a = v² / r

let's calculate

            a = 28² / 15.0

            a = 52.26 m / s²

as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards

Point B

           a ’= 142/15

           a ’= 13.06 m / s²

in this case the acceleration is vertical downwards

c) The values ​​of the normal force

point A

let's use Newton's second law

           ∑ F = m a

           N- W = m a

           N = mg + ma

           N = m (g + a)

           N = 450.0 (9.8 + 52.25)

           N = 2.79 10⁴ N

d) Point B

            -N -W = m (-a)

             N = ma -m g

             N = m (a-g)

             N = 450.0 (14.0 - 9.8)

             N = 1.89 10³ N

Two objects are electrically charged. The net charge on one object is doubled.

Therefore, the electric force _____.

reverses

doubles

quadruples

divides

Answers

Therefore, the electric force reverses.

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

.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

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

A rocket, with a mass of 5100 kg, has an engine that provides a net upward force of 8.0 x 10^5 N. It starts from rest and reaches a maximum speed of 900 m/s. How long does it take to reach that maximum velocity?

Answers

Answer:

5.7375 seconds

Explanation:

The computation of the time required to reach that maximum velocity is shown below:

Given that

Mass = m =  5100 kg

Net upward force F = 8 × times 10^5 N

Initial speed = V_i = 0

Maximum speed = V = 900 m.s

Based on the above information

Impluse J = m(V - V_i)

= 5100 (900 - 0)

= 459 × 10^4 kg m.s

As we know that

J = FT

So

T = J ÷ F

= (459 × 10^4) ÷ (8 × 10^5)

= 5.7375 seconds

QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!

Answers

Answer:

A. v = √2gh

B. No! The final velocity does not depend on the mass of the car.

C. Yes! the final velocity depends on the steepness of the hill

D. 3.28 m/s

Explanation:

A. Determination of the final velocity.

½mv² = mgh

Cancel out m

½v² = gh

Cross multiply

v² = 2gh

Take the square root of both side

v = √2gh

B. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is no mass (m) in the formula.

Thus, the final velocity does not depend on the mass of the car.

C. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is height (h) in the formula.

Thus, the final velocity depends on the steepness of the hill

D. Determination of the final velocity.

Height (h) = 0.55 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = √2gh

v = √(2 × 9.8 × 0.55)

v = √10.78

v = 3.28 m/s

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 .

Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant conditioning B. determinism C. self-efficacy D. self-worth

Answers

Answer:

C

Explanation:

Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.  

What is Self efficacy?

A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).

The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.

Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.

Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.  

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The sides of a right triangle that has any given vector for the hypotenuse are called _____
A. Scalar
B. Component
C. Resultant
D. Vector

Answers

Answer:

They are the resultant vector.

A triangle has the resultant vector

Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.

Answers

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

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

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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"

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

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

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.

because there the sperm and eggs are combining together to produce
so thats why they look alike

Answers

Answer:

yes. that is how a baby is conceived.

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

A bucket is filled partly with water such that its combined mass is 2.17 kg. It is tied to a rope and whirled in a circle with a radius of 1.13 m. The speed at the top of the circle is 4.42 m/s and the speed at the bottom of the circle is 7.1 m/s. (a) Determine the acceleration, net force and tension force at the top of the circle.

Answers

Answer:

1) a = 17.3 m/s²

2) Fnet = 37.5 N

3) T = 16.2 N

Explanation:

1)

When the bucket is at the top of the circle, there are two forces acting on it: the tension force (T) which pulls from the bucket, so it is directed downward, and the force due to gravity, that also points downward, so both forces add:[tex]F_{net} = T + m*g (1)[/tex]According to Newton's 2nd Law, this net force must be equal to the mass of the bucket, times the acceleration.Now, due to the bucket is moving around a circle, there must be a force that keeps the bucket following a circular trajectory, that is the centripetal force, and always aims toward the center of the circle.This force is not a new type of force, it's always the net force that aims toward the center.At the top of the circle, because as the tension force as gravity point downward, the centripetal force, is just this net force.It can be showed that the centripetal force can be written as follows:

      [tex]F_{c} = m*a_{c} = m*\frac{v^{2}}{r} (2)[/tex]

Since we have already said that  a = ac (At the top of the circle), we can solve (1) for a, simplifying and replacing v and r by their values, as follows:

       [tex]a = a_{c} = \frac{v^{2} }{r} = \frac{(4.42m/s)^{2} }{1.13m} = 17.3 m/s2 (3)[/tex]

2)

Once we got the value of a, applying Newton's 2nd law, we can find easily the net force on the bucket at the top of the circle, as follows:

       [tex]F_{net} = m*a = 2.17 kg * 17.3 m/s2 = 37.5 N (4)[/tex]

3)

We have already said, that at the top of the circle, the net force is just the sum of the tension T and the force of gravity, as follows:

       [tex]F_{net} = T + m*g = 37.5 N (5)[/tex]

Replacing m and g by their values, we can solve (4) for T:

       [tex]T = 37.5 N - m*g = 37.5 N - (2.17kg*98m/s2) \\ = 37.5 N - 21.3 N = 16.2 N (6)[/tex]

Physical quantities are generally not purely numerical: They have a particular dimension or combination of dimensions associated with them. Thus, your height is not 74, but rather 74 inches, often expressed as 6 feet 2 inches. Although feet and inches are different units they have the same dimension--length.

Required:
a. Find the dimensions [V] of volume.
b. Find the dimensions [v] of speed.

Answers

Answer:

a, The dimension of volume is L³

b. The dimension of speed = L/T

Explanation:

The three fundamental quantities are Mass, Length and time. Other physical quantities are obtained from or derived from these three. These other quantities are known as derived quantities. The units of the fundamental quantities are Kilogram (kg) for Mass, meters for length, and second for time.

In the given question:

a. Volume = length * breadth * height

since breadth and height all measure length, the dimension of volume becomes:

volume = length * length * length = L³

Thus, the dimension of volume is L³

b. Speed, v = distance/time

Distance measures length, therefore, the dimension of speed will be:

Speed = length / time = L/T

Therefore, the dimension of speed = L/T

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:

Cameron is standing on the edge of a 60 m high cliff. He horizontally throws a football
with an initial velocity of 10 m/s. How long does it take for the football to hit the
ground?

Answers

Answer:34.6 m/s

Explanation: It is asking how long meaning the answer is in time

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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