Use the following information for questions 18 - 21. A 0.13 kg disk is rotating at an angular speed of 57 rad/s. The disk has a radius of 0.25 m. The disk speeds up for 3 s. After the 3 s have passed, the edge of the disk is under a centripetal force of 312.13 N. What is the centripetal acceleration of the disk at this time

Answers

Answer 1

Answer:

[tex]a=2401m/s^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=0.13kg[/tex]

Angular speed [tex]\omega=57rads/s[/tex]

Radius [tex]r=0.25m[/tex]

Time [tex]t=3s[/tex]

[tex]F=312.13N[/tex]

Generally the equation for centripetal acceleration is mathematically given by

[tex]a=\frac{f}{m}[/tex]

[tex]a=\frac{312.13}{0.13}[/tex]

[tex]a=2401m/s^2[/tex]


Related Questions

According to the model, when was the universe at its most dense?

A) During the Dark Ages where matter increased in mass.

B) Just before the Big Bang where all matter existed in a singularity.

C) During the nuclear fusion events, as the atoms become more massive.

D) Current day, as the number of galaxies, solar systems, and planets have increased.

Answers

it d cus you look at the gragh

Answer:

The Answer is D

Explanation:

Hope this helps!!!!

A ship is flying away from Earth at 0.9c (where c is the speed of light). A missile is fired that moves toward the Earth at a speed of 0.5c relative to the ship. How fast does the missile move relative to the Earth

Answers

Answer:

the required speed with which the missile move relative to the Earth is -0.727c

Explanation:

Given the data in the question;

relative velocity relation;

u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]

so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]

and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]

hence, u' is the speed of B as seen by A

so

u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]

now, given that; V[tex]_A[/tex] = 0.9c  and V[tex]_B[/tex]  = 0.5c

we substitute

u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]

u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]

u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)

u' = ( -0.4c ) / 1 - 0.45

u' = -0.4c / 0.55

u' = -0.727c

Therefore, the required speed with which the missile move relative to the Earth is -0.727c

A scenario where reaction time is important is when driving on the highway. During the delay between seeing an obstacle and reacting to avoid it (or to slam on the brakes!) you are still moving at full highway speed. Calculate how much distance you cover in meters before you start to put your foot on the brakes if you are travelling 65 miles per hour.

Answers

Answer:

66.83 meters

Explanation:

After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.

1 mile = 1609.34 meters  (multiply these meters by 65)

65 miles = 104,607 meters

1 hour = 3600 seconds

Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.

[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]

Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...

29.0575m * 2.3s = 66.83 meters

As shown in Fig. 4, an ideal gas of monatomic molecules expands from its initial state A to a state B through an isobaric process and then further expands to a volume C. Find the work done by the gas, increase in internal energy, and the energy transferred by heat to the gas over the whole process​

Answers

The picture is very blurry

When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion.

Answers

Answer:

[tex]F=3139.2N[/tex]

Explanation:

Mass [tex]m=40.0kg[/tex]

Acceleration [tex]a=7g=68.67m/s^2[/tex]

With g as 9.81

Generally the equation for Force is mathematically given by

  [tex]F_net=F-w\\\\F-w=ma\\\\F=ma-w\\\\[/tex]

 [tex]F=ma-mg\\\\F=m(a+g)[/tex]

Therefore

 [tex]F=m*8g[/tex]

 [tex]F=40*8*9.81[/tex]

 [tex]F=3139.2N[/tex]

Object A is 8 kg and at rest and Object B is 16 kg and moving at 10 m/s to the Right. If Object B hits Object A and Object B is at rest after the collision, what is the velocity of Object A after the collision? Why?

Answers

Answer:

Initially...

Object A is at rest(v=0)

With Mass =8kg

Object B moving at 10ms- and Mass of 16kg.

Now After collision(Elastic)... B Came to rest...Meaning its final vel =0

But its mass remains the same

What velocity would A be moving with?

From Conversation of Momentum

M'U' + M"U"= M'V' + M"V"

8(0) + (16)(10) = 8V' + 16(0)

0 + 160 = 8V' + 0

8V'= 160

V'= 160/8

V'=20ms-¹.

A would be Moving at 20ms-¹ after collision.

What is energy?
Select one:

A force that must be exerted in order to accelerate an object.

A property that must be transferred to perform work.

A particle that can be absorbed to change momentum.

A system that can be rearranged in order to change its state.

Answers

Answer:

Energy is:

A property that must be transferred to perform work.

Answer:

Energy

Explanation:

A property that must be transferred to perform work .

Hope it is helpful to you Stay safe healthy and happy ☺️

The lever of a car lift has an area of 0.2 meters squared, and the area of the lift under the car is 8
meters squared. If you push with a force of 3 newtons, how much force will be applied to the
car?

Answers

Answer:

THE ANSWER IS SOMETHING LIKE 55

eididjsmsisijsjsiakaksannnahshsjeejekekekkeie​

Answers

Answer:

ok

Explanation:

nskakkskdnsksmskzkksnsnxksjsjos

Which option is the best blackbody radiator?

A.
The Sun

B.
A red laser pointer

C.
A tennis ball

D.
Boiling water

Answers

Answer:

A. The Sun

Explanation:

The Sun is to be considered a perfect black body.

What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?​

Answers

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

A car moving in a straight line uniformly accelerated speed increased from 3 m / s to 9 m / s in 6 seconds. With what acceleration did the car move?


a.
2 m/s2


b.
1 m/s2


c.
0 m/s2


d.
3 m/s2

Answers

I think it’s 1 . The answer is B

Answer:

b) 1 m/s

I am sure...........

If the pressure of a gas is really due to the random collisions of molecules with the walls of the container, why do pressure gauges – even very sensitive ones – give perfectly steady readings? Shouldn't the gauge be continually jiggling and fluctuating? Explain.

Answers

Answer:

there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

Explanation:

The pressure measurement is carried out by calibrating the force exerted by the air on a surface of known area, suppose a small area 1 mm² = 0.01 cm²

To find out if the random movement of air molecules affects the pressure reading, let's calculate the number of molecules that reaches the pressure gauge.

In a system at atmospheric pressure and in a volume of 1 m³ (walls of 1 m each) there is one mole of air molecules, this mole is evenly distributed, so how many molecules fall on our surface

           # _molecule = 6.02 10²³ 0.01 10⁻⁴ / 1

           #_molecular = 6.02 10¹⁷ molecules per second

therefore the variation of the number of molecules is not very important

Consequently there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.

The latent heat of vaporization of water is roughly 10 times the latent heat of fusion of water. The amount of heat required to boil away 1 kg of water is __________ the amount of heat required to melt 1 kg of ice.

Answers

Answer:

The amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice

Explanation:

let the latent heat of fusion of ice  = L

then, the latent heat of vaporization of water = 10L

The heat of fusion of 1 kg of ice = 1 x L = L

The heat of vaporization 1 kg of water = 1 x 10L = 10L

Therefore, the amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice

A car moves at a constant speed of 90km/h from a starting point. Another car moves at 70km/h after 2hours from the same starting point. if both cars moves in the same direction, after how many hours will the distance between the first car and the second car to be 40 km.​

Answers

Answer:

400

Explanation:

Un recipiente contiene 224 dm3 de Ozono de masa 4.561 Kg a 51.09 grados celsius. Calcula la presión del Ozono

Answers

Answer:

Por lo tanto, la presión del ozono es:

[tex]P=0.011\: atm[/tex]  

Explanation:

Podemos usar la ecuacion de los gases ideales;

[tex]PV=nRT[/tex] (1)

Tenemos:

El volumen V = 224 dm³ = 224 L

La temperatura T = 51.09 C = 324.09 K

La masa es m = 4.561 kg

Lo necesitamos ahora es calvular n que es el numero de moles;

recordemos que el peso molecular del ozono M = 48 g/mol.

[tex]n=\frac{m}{M}=\frac{4.561}{48}=0.095\: mol[/tex]

Finalmente, usando la ecuacion 1 despejamos la presion P

[tex]P=\frac{nRT}{V}[/tex]

[tex]P=\frac{0.095*0.082*324.09}{224}[/tex]  

Por lo tanto, la presion del ozono es:

[tex]P=0.011\: atm[/tex]  

Espero te haya ayudado!

Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m

Answers

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

[tex]T = 2 \pi \sqrt\frac{m}{k}[/tex]

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

[tex]T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s[/tex]

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

If 1.02 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved through it?

Answers

Answer:

Explanation:

one electron has [tex]1.60217662*10^{-19}~coulombs~then\\\\1.02*10^{20}~electrons------->1.02*10^{20}*1.60217662*10^{-19}~coulombs= 16.3422~coulombs[/tex]

how many grams are in 5 kilos​

Answers

1 kilo = 1000 gms

Therefore, 5 kilo = 5000 gms

There are 5,000 grams in 5 kilos

(1) Define uniform acceleration

Answers

Answer:

Explanation:

When an object's speed increases at a constant rate, we say it has constant/uniform acceleration.

Calculate the equivalent resistance

Answers

Not enough information

15. A car travelling towards the right has a mass of 1332 kg and has a speed of 25 m/s. A truck is
travelling towards the left with a mass of 3000 kg and a speed of 15 m/s. They collide head
on with each other. What is the total momentum after the crash? In which direction will the
vehicles travel after the collision?

Answers

Explanation:

Given that,

The mass of a car, m₁ = 1332 kg

The speed of the car, u₁ = 25 m/s (right)

The mass of a truck, m₂ = 3000 kg

The speed of the truck, u₂ = -15 m/s

The total momentum after the crash is given by :

p=m₁u₁ + m₂u₂

Put all the values,

P = 1332(25) + 3000(-15)

= −11700 kg-m/s

So, the total momentum after the crash is equal to 11700 kg-m/s and it is in the left direction.

Select the correct answer from each drop-down menu.
Balloon B is negatively charged.
Balloon A is ( )
charged, and balloon C is( ) charged. If balloon A approaches balloon C, there'll be a force of ( )
between them.

Answers

Answer:

it would look like what I did in the picture

Explanation:

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of the earth. The second satellite travels at aspeed that is less than v.At what distance from the center of the earth does the secondsatellite orbit?At a distance that is less than r.At a distance equal to r.At a distance greater than r.Now assume that a satellite of mass m is orbiting the earth at a distance r from the center of the earth with speed v_e. An identical satellite is orbiting the moon at thesame distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make onerevolution compare to the time T_e it takes the satellite orbiting the earth to make onerevolution?T_m is less than T_e.T_m is equal to T_e.T_m is greater than T_e.

Answers

Answer:

a. At a distance greater than r

b. T_m is greater than T_e.

Explanation:

a. Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed vat a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?

Since the centripetal force on any satellite, F equals the gravitational force F' at r,

and F = mv²/r and F' = GMm/r² where m = mass of satellite, v = speed of satellite, G = universal gravitational constant, M = mass of earth and r = distance of satellite from center of earth.

Now, F = F'

mv²/r = GMm/r²

v² = GM/r

v = √GM/r

Since G and M are constant,

v ∝ 1/√r

So, if the speed decreases, the radius of the orbit increases.

Since the second satellite travels at a speed less than v, its radius, r increases since v ∝ 1/√r.

So, the distance the second satellite orbits is at a distance greater than r

b. An identical satellite is orbiting the moon at the same distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make one revolution compare to the time T_e it takes the satellite orbiting the earth to make one revolution?

Since the speed of the satellite, v = √GM/r where M = mass of planet

Since the satellite is orbiting at the same distance, r is constant

So, v ∝ √M

Since mass of earth M' is greater than mass of moon, M", the speed of satellite circling moon, v_m is less than v the speed of satellite circling earth at the same distance, r

Now, period T = 2πr/v where r = radius of orbit and v = speed of satellite

Since r is constant for both orbits, T ∝ 1/v

Now, since the speed of the speed of the satellite on earth orbit v  is greater than the speed of the satellite orbiting the moon, v_m, and T ∝ 1/v, it implies that the period of the satellite orbiting the earth, T_e is less than the period of the satellite orbiting the moon, T_m since there is an inverse relationship between T and v. T_e is less T_m implies T_m is greater than T_e

So, T_m is greater than T_e.

why the walls of tyres becomes warm as the car moves​

Answers

Answer:

the particles vibrate inside the tyre

Explanation:

as the car moves kinetic energy is transfered in the tyres which causes the particles to vibrate inside the tyre so the kinetic store is. transferred into thermal

can Denel be regarded as monopoly in south africa​

Answers

Answer:

Denel (Pty) Ltd was established as a private company, incorporated in terms of the Companies Act on 1 April 1992 with the State as the sole shareholder. ... Denel can at present, without doubt, be regarded as a public monopoly.

why are cows is important?​

Answers

Answer:

cause they give u milk

Explanation:

Answer:

Cows are important as they provide humans many things for survival. They provide milk, meat, and leather, all of these are important resources.

A string that is under 50.0N of tension has linear density 5.0g/m. A sinusoidal wave with amplitude 3.0cm and wavelength 2.0m travels along the string. What is the maximum speed of a particle on the string

Answers

Answer:

9.42 m/s

Explanation:

Applying,

V' = Aω.............. Equation 1

Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.

But,

ω = 2πf................. Equation 2

Where f = frequency, π = pie

And,

f = v/λ................ Equation 3

Where, λ = wave length, v = velocity

Also,

v = √(T/μ)................. Equation 4

Where T = Tension, μ = linear density.

From the question,

Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m

Substitute into equation 4

v = √(50/0.005)

v = √(10000)

v = 100 m/s

Also Given: λ = 2.0 m

Substitute into equation 3

f = 100/2

f = 50 Hz.

Substitute the value of f into equation 2

Where π = constant = 3.14

ω = 2(3.14)(50)

ω = 314 rad/s

Finally,

Given: A = 3.0 cm = 0.03 m

Substitute into equation 1

V' = 0.03(314)

V' = 9.42 m/s

The efficiency of a machine can be increased by

Answers

Explanation:

the efficiency of a machine can be increased by reducing the friction

please mark the brainliest

Reducing the friction

N 4. Which of the following can cause a short circuit?

Answers

Answer:

pretty sure its A

Explanation:

please give brainliest if i'm correct

Answer:

A.

Explanation:

I have done this and that was correct

hope this helps

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