A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop. (a) Assuming a constant angular acceleration, find the time for it to come to rest. (b) What is its angular acceleration

Answers

Answer 1

Answer:

The time of motion is 333.3 s

The angular acceleration is -0.0045 rad/s²

Explanation:

Given;

angular distance of the flywheel, θ = 40 rev

initial angular speed, [tex]\omega_i[/tex] = 1.5 rad/s

When the wheel comes to rest, the final angular speed, [tex]\omega_f[/tex] = 0

The angular acceleration is calculated as follows;

[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta \\\\0 = (1.5 \ rad/s)^2 + 2\alpha (40 \ rev\times \frac{2\pi \ rad}{1 \ rev} )\\\\0 = 2.25 + 160\pi \alpha\\\\160\pi \alpha = - 2.25\\\\\alpha = -\frac{2.25 }{160\pi} \\\\\alpha = -0.0045 \ rad/s^2[/tex]

The time of motion is calculated as;

[tex]\omega_f = \omega _i + \alpha t\\\\0 = 1.5 + (-0.0045t)\\\\0 = 1.5 - 0.0045t\\\\0.0045t = 1.5\\\\t = \frac{1.5}{0.0045} = 333.3 \ s[/tex]


Related Questions

A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the mass of the body

Answers

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, [tex]\gamma _0[/tex] = 0.875

mass of the object in oil, [tex]M_o[/tex] = 0.013 kg

mass of the object in water, [tex]M_w[/tex] = 0.012 kg

let the mass of the object in air = [tex]M_a[/tex]

weight of the oil, [tex]W_0 = M_a - 0.013[/tex]

weight of the water, [tex]W_w = M_a - 0.012[/tex]

The relative density of the oil is given as;

[tex]\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg[/tex]

Therefore, the mass of the body is 0.02 kg.

Why are the largest craters we find on the Moon and Mercury so much larger than the largest craters we find on the Earth

Answers

Answer:

Because Moon and Mars has no atmosphere.

Explanation:

Moon and Mars has no atmosphere, so there is no friction on the falling object due to the atmosphere. The speed of the falling object is more at Moon and Mars.

When a small object impact on the surface of moon or Mars with high speed, the size of crater is large than the earth as out earth has atmosphere.

what is gama rays an it's uses​

Answers

Answer:

GAMMA RAYS:

A photon emitted spontaneously by a radioactive substance also : a photon of higher energy than that of an X-ray.

USES OF GAMMA RAYS:

Gamma rays are used in medicine (radiotherapy), industry (sterilization and disinfection), and the nuclear industry. Shielding against gamma rays is essential because they can cause diseases to skin or blood, eye disorders, and cancers.

A ball of mass 0.3 kg is released from rest at a height of 8 m. How fast is it going when it hits the ground? (Gravity being equal to 9.8)

Answers

Answer:

Explanation:

Mass doesn't matter here because when something is falling, gravity plays fairly; an elephant falls at the same rate of acceleration as does a feather. What DOES matter is everything pertinent to the y-dimension of free-fall:

a = -9.8 m/s/s

v₀ = 0 (since the ball was held before it was dropped)

v = ??

Δx = -8 m (negative because the ball drops this far below the point from which it was released).

Putting all this together in one equation:

v² = v₀² + 2aΔx and filling in this equation:

v² = (0)² + 2(-9.8)(-8) and

v² = 156.8 so

v = 12.5 which rounds to 13 if you're using 2 sig figs, and rounds to 10 if you're only using 1 (which you should be, according to the way the numbers have been given in this problem)

The work function for silver is 4.73 eV. (a) Convert the value of the work function from electron volts to joules.

Answers

Answer:

[tex]W=7.56\times 10^{-19}\ J[/tex]

Explanation:

Given that,

The work function for silver is 4.73 eV.

We need to find the value of the work function from electron volts to joules.

We know that,

[tex]1\ eV=1.6\times 10^{-19}\ J[/tex]

For 4.73 eV,

[tex]4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J[/tex]

So, the work function for silver is [tex]7.56\times 10^{-19}\ J[/tex].

1. 20kg of water is ejected horizontally in 10s; the speed of the water leaving the nozzle is 30m/s. Calculate the force experienced by a fire-fighter holding the hose. ​

Answers

Answer 60 NEWTON

Explanation:

FORCE = MASS * acceleration

acceleration= VELOCITY / TIME

acceleration= 30 / 10 = 3   M/S2

FORCE = MASS * acceleration

FORCE = 20 *3 = 60 NEWTON

While using a digital radiography system, suppose a radiographer uses exposure factors of 10 mAs and 70 kVp with an 8:1 grid for an AP shoulder radiograph with acceptable anatomical part penetration and detector element (DEL) exposure. If the radiographer desires to increase scatter absorption using a 12:1 grid, what new exposure factors should be used to maintain the same DEL exposure

Answers

Answer:

b. 12.5 mAs, 70 kVp

Explanation:

The given parameter are;

The initial exposure factors := 10 mAs and 70 kVp

The initial Grid Ratio, G.R.₁ = 8:1

The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂  = 12:1

Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;

The GCF for G.R. 8:1 = 4

The GCF for G.R. 12:1 = 5

Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;

mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs

Therefore the new exposure factors are;

12.5 mAs, 70 kVp

Quickly pls!!! A wave with a wavelength of 0.5 m moves with a speed of 1.5 m/s. What is the frequency of the wave?
A. 2.0 Hz
B. 1.0 Hz
C. 0.33 Hz
D. 3.0 Hz

Answers

Speed/wavelength - 1.5m/s / 0.5 m
The frequency of the wave will be 3 or D.

100 POINTS !!! PLEASE HELP !!!!
What is the independent variable?
What is the dependent variable?
I
Materials

1. 4 antacid tablets
2. 2 clear cups labeled Hot and Cold
3. Water
4. Timing device

Answers

Answer:

dependent: the outcome of the experience

independent variable: everything literaly.

Independent is where you change some variables and see the result

Dependent is literaly the result, or the outcome dependent on the exprience.

Explanation:

I got u.

a sprinter runs 100m in12.5seconds calculate the speed

Answers

Answer:

8m in 1 second

480m = 1 hour

11,520m in one day

Explanation:

Hope this is helpful

the bodies in this universe attract one another name the scientist who propounded this statement​

Answers

Answer:

It was proposed by Isaac Newton

Explanation:

The law of universal attraction of expression

          F = [tex]G \ \frac{m_1m_2}{ r^2}[/tex]G m1m2 / r ^ 2

where G is a constant, m₁ and m₂ are the masses of the bodies and r the distance between them.

It was proposed by Isaac Newton

With this law Newton explained that the force that pulls the moon towards the earth is the same as that which attracts an apple towards the earth

difine scalar quantity​

Answers

Scalar quantity are physical quantities that have just magnitude, not direction.

It is always positive.Examples: Speed, distance

Light strikes a smooth wooden tabletop.

What happens to the light after it is reflected?

The light rays bounce off the table and all move in the same direction.
The light rays bounce off the table and move in different directions.
The light rays pass through the table and all move in the same direction.
The light rays pass through the table and move in different directions.

Answers

Answer:

For smooth surface:The light rays bounce off the table and all move in the same direction.

The light will bounce off and continue in the direction it bounced.

An ammeter with a resistance of 5.0 ohm is connected in series with a 3.0V cell and a lamp rated at 300 mA, 3V. Calculate the current that the ammeter will measure

Answers

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

. A car accelerates uniformly from 0 to 72 km/h in 11.5 seconds.
a) What is the acceleration of the car in m/s2? b) What is the position of the car by the time it reaches the velocity of 72 km/h?​

Answers

Answer:

Explanation:

First job is to convert 72 km/hr to m/s:

[tex]72\frac{km}{hr}[/tex] × [tex]\frac{1000m}{1km}[/tex] × [tex]\frac{1hr}{3600s}[/tex] = 2.0 × 10¹ m/s

Now to find the acceleration which is

[tex]a=\frac{v_f-v_0}{t}[/tex] and filling in:

[tex]a=\frac{2.0*10^1-0}{11.5}=1.7\frac{m}{s^2}[/tex] That's part a. Part b want to know how far the car can get in 11.5 seconds (because that's the time it takes for the car to get to 72 km/hr). Since we know that the car can get 2.0 × 10¹ meters in 1 second, that means that in 11.5 seconds, the car can get 11.5(2.0 × 10¹) which is 230 meters.

A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total) displacement). Round your answer to the appropriate significant figures. Round your angle to the nearest degree.

Answers

In component form, the displacement vectors become

• 350 m [S]   ==>   (0, -350) m

• 400 m [E 20° N]   ==>   (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W]   ==>   (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

how long will it take a car to go from 0 to 45 km/hr if they are accelerating at 5 km/hr/s?

Answers

Answer:

9 seconds

Explanation:

[tex]acceleration = \frac{final \: speed - initial \: speed}{time \: taken} [/tex]

[tex]5km/ hr/ s = \frac{45 - 0 (km/ h)}{t} [/tex]

cross multiplying

[tex]5t = 45 secs \\ t = 9 \: secs[/tex]

it will take 9secs for the car to go from 0 to 45 km/ h

A system has a pressure of 5 N/m2
If a force of 2000N is applied, what is the area that the force is applied to?
Give the units.

Answers

400 m2

Explanation:

Pressure = Force ÷ Area

5 N/m2 = 2000 N ÷ A

A = 2000 N ÷ 5

= 400 m2

If a force of 2000N is applied, the area that the force is applied to is 400 m²

What is force?

The word "force" has a specific meaning in science. At this level, calling a force a push or a pull is entirely appropriate. A force is not something an object "has in it" or that it "contains."

One thing experiences a force from another. There are both living things and non-living objects in the concept of a force.

The amount of force applied to a certain region is referred to as pressure. The force per unit area is called pressure. F in this condensed version of the equation stands in for the force, which is expressed in newtons.

Given that the pressure of 5 N/m²

Force is 2000N

Pressure = Force ÷ Area

5 N/m² = 2000 N ÷ A

A = 2000 N ÷ 5 = 400 m²

Therefore, the area that the force is applied to is 400 m².

To learn more about force, refer to the link:

https://brainly.com/question/19529052

#SPJ2

Calculate the distance travelled by the car in part Q use the equation distance travelled= average speed x time

Answers

Explanation:

distance travelled = average speed x time

                               =30m/s*100s

                                =3000m

Answer:

3000m

Explanation:

30m/s*100s

3000m

Classes. frequency
0-20 2
20-40. 2
40-60. 3
60-80 12
80-100 18
100-120 5
120-140. 2

Find mean, median and mode​

Answers

Answer:

The mean is 79.[tex]\overline {54}[/tex]

The median is 80 - 100

The mode is 80 - 100

Explanation:

The given table is presented as follows;

[tex]\begin{array}{lcrc}Classes&Mid \ point &Frequency &Frequency \times Mid \ point\\0 - 20&10& 2&20\\20-40&30&2&60\\40-60&50&3&150\\60-80&70&12&840\\80-100&90&18&1620\\100-120&110&5&550\\120-140&130&2&260\end{array}[/tex]

The mean of a class of values, [tex]\overline x[/tex] = ∑(Frequency × Midpoint)/∑(Frequency)

Therefore, we get;

[tex]\overline x[/tex] = (20+60+150+840+1620+550+260)/(2+2+3+12+18+5+2) = 79.[tex]\overline {54}[/tex]

The mean, [tex]\overline x[/tex] =79.[tex]\overline {54}[/tex]

The median class = The middle value lass = The class at the 22 nd value = 80 - 100

The median = 80 - 100

The modal class = The class with the highest frequency = 80 - 100

The mode = 80 - 100

Internal energy of a diatomic gas consists of:
OA. kinetic energy due to vibration and rotation.
B. kinetic energy due to translation, vibration, and rotation.
C. potential energy due to intermolecular forces.
D. kinetic energy due to translation only.

Answers

Answer:

C) Potential energy due to intermolecular forces.

As a main sequence star exhausts hydrogen in its core, its surface becomes ___ and its energy output per second (luminosity) becomes ____.

Answers

Answer:

b

Explanation:

bc

Suponga que la pelota de la figura se proyecta desde una altura de 35.0 m sobre el suelo y se le imprime una velocidad horizontal inicial de 8.25 m/s. a) ¿Cuánto tiempo tardará la pelota en golpear el suelo? b) ¿A qué distancia del edificio tocará el suelo la pelota?

Answers

Answer:

Explanation:

qoe39rwecwurbq;7edfqbsi

The Displacement is 5m. We found that using the
Pythagorean Theorem.
Vector Quantities require both a Displacement and a
Direction.
What direction is this Vector?
South
Northeast
West

Answers

Answer:

A vector can be written as:

(R, θ)

Where R is the magnitude, in this case, we know that the magnitude of the displacement is 5m

Then:

R = 5m

and θ defines the direction, it's an angle measured from the positive x-axis.

(In the image, θ would be the angle located at the point A)

Now, if you look at the image, you can see a triangle rectangle.

Where the adjacent cathetus has a length of 4,

the opposite cathetus has a length of 3 units

the hypotenuse has a length of 5 units.

So we can use any trigonometric rule to find the value of θ, like:

sin(θ) = (opposite cathetus)/hypotenuse

Then:

sin(θ) = 3m/5m

Now we can use the inverse sin function, Asin(x), in both sides

Asin( sin(θ)) = θ = Asin( 3/5) = 36.87°

then the vector is:

(5m, 36.87°)

Now, if we define the positive y-axis as the North, and the positive x-axis as the East.

This vector would point at 36.87° North of East.

(or almost Northeast)

A 5.0 kg block of ice is at rest at the top of a smooth inclined plane. The block is released and slides 2.0 m down the plane. Assuming there is no friction between the block and the surface, calculate
a) the gravitational potential energy at the top of the plane
b) the component of the weight parallel to the plane
c) the acceleration of the block
d) the velocity of the block at the bottom of the plane
e) the kinetic energy at the bottom of the plane. ​

Answers

Answer:

a) 98.1 Joules

b) 49.05 N × sin(θ)

c) 9.81 × sin(θ)

d) The velocity of the block at the bottom of the plane, v is approximately 6.264 m/s

e) 98.1 Joules

Explanation:

The given parameters of the block are;

The mass of the block, m = 5.0 kg

The distance down the plane the block slides, h = 2.0 m

The friction between the block and the surface = 0

Let θ represent the angle of inclination oof the plane

a) The gravitational potential energy, P.E. = m·g·h

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ P.E. ≈ 5.0 kg × 9.81 m/s² × 2.0 m = 98.1 Joules

The gravitational potential energy, P.E. ≈ 98.1 Joules

b) The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex], is given as follows;

[tex]w_{\parallel}[/tex] = w × sin(θ) = m·g·sin(θ)

∴ [tex]w_{\parallel}[/tex] ≈ 5.0 kg × 9.81 m/s² × sin(θ) = 49.05 × sin(θ) N

The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex] ≈ 49.05 N × sin(θ)

c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;

[tex]w_{\parallel}[/tex] = m·g·sin(θ) = m·a

Where a represents the acceleration of the block along the plane

Therefore, by comparison, we have;

g·sin(θ) = a

∴ a ≈ 9.81 × sin(θ)

d) Given that the motion of the block is 2.0 m downwards, we have;

The velocity of the block at the bottom of the plane, v² = 2·g·h

Therefore, v² ≈ 2 × 9.81 m/s²× 2.0 m = 39.24 m²/s²

v = √(39.24 m²/s²) ≈ 6.264 m/s

e) The kinetic energy at the bottom of the plane, K.E. = (1/2)·m·v²

∴ K.E. = (1/2) × 5.0 kg × 39.24 m²/s² = 98.1 J

suppose the pilot starting again from rest opens the throttle part.way at constant acceleration the airboat then covers a distance of 60.0m in 10.0s find the net force action on the boat​

Answers

Answer:

Acceleration is 1.2 m/s^2.

Explanation:

initial velocity, u = 0

distance, d = 60 m

time, t = 10 s  

Let the acceleration is a.

use second equation of motion

[tex]s= u t +0.5 at^2\\\\60 = 0 + 0.5 \times a \times 10\times 10\\\\a = 1.2 m/s^2[/tex]

Now according to the Newton's second law

Force = mass x acceleration

Let the mass is m.

F = m x 1.2 = 1.2 m Newton  

A 72 kg swimmer dives horizontally off a raft floating in a lake. The diver's speed immediately after leaving the raft is 3.8 m/s. If the time interval of the interaction between the diver and the raft is 0.25 s, what is the magnitude of the average horizontal force by diver on the raft?

Answers

Answer:

F = 1094.4 N

Explanation:

From impulse - momentum theorem, we now that ;

Impulse = momentum

Where;

Formula for impulse = force (F) × time(t)

Momentum = mass(m) × velocity(v)

Now, we are given;

Mass of swimmer; m = 72 kg

Speed; v = 3.8 m/s

Time; t = 0.25 s

Thus;

F × t = mv

F = mv/t

F = (72 × 3.8)/0.25

F = 1094.4 N

This value of force is the magnitude of the average horizontal force by diver on the raft.

Beginning at the synapse of a neuron, place the events in neuronal signaling in the correct sequence.1. Ion channels bind the ligand and open.2. Na+ is pumped out of the cell, and the membrane potential is restored.3. Acetylcholinesterase breaks down acetylcholine.4. Na+ enters the postsynaptic cell, and the membrane potential changes.5. Acetylcholine is released into the synapse.6. Na+ ion channels close.A. 3, 4, 6, 1, 5, 2B. 5, 1, 4, 3, 6, 2C. 4, 5, 1, 2, 6, 3D. 1, 2, 3, 4, 5, 6E. 6, 4, 3, 1, 2, 5

Answers

Answer:

B. 5, 1, 4, 3, 6, 2

Explanation:

Acetylcholine (ACh) is a molecule that functions as a neurotransmitter and plays diverse roles in brain and muscle function. Ach is released by the presynaptic neurons into the synaptic cleft. The nicotinic acetylcholine receptors are ACh ligand-gated ion channels located at the postsynaptic membrane which admit both sodium (Na+) and potassium (K+) ions. These receptors bind ACh, increasing the permeability of the postsynaptic membrane to Na+ and K+ ions. In consequence, ion channels open and sodium (Na+) ions enter the postsynaptic cell, causing the postsynaptic membrane to depolarize. An enzyme known as acetylcholinesterase breaks down acetylcholine into acetate and choline. If depolarization passes a threshold potential, an action potential is generated in the postsynaptic membrane. The generation of a postsynaptic potential at the excitatory synapse is known as excitatory postsynaptic potential. Once depolarization is complete, Na+ channels close and membrane permeability to Na+ ions declines to the resting levels.

A jet plane lands at a speed of 100 m/s and can accelerate at a maximum rate of -5.00 m/s^2 as it comes to a rest.
(a from the instant the plane touches the runaway, what is the minimum time needed before it can come to a rest?
(b Can this plane land on a runaway that is only 0.800 km long?
shown work pls will reward alot of points

Answers

Answer:

a)   t = 20 s,  b)  x = 1000 m, As the runway is only 800 m long, the plane cannot land at this distance

Explanation:

This is a kinematics exercise

a) in minimum time to stop,

           v = vo + at

           v = 0

           t = -v0 / a

we calculate

          t = -100 / (5.00)

          t = 20 s

b) Let's find the length you need to stop

          v² = vo² + 2 a x

          x = -v0 ^ 2 / 2a

          x = - 100² / 2 (-5.00)

          x = 1000 m

As the runway is only 800 m long, the plane cannot land at this distance.

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m each. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s2 during

Answers

Complete question is;

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m each. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s² during one of the running portions, what is her final velocity at the end of the 100.0 m? Round your answer to the nearest tenth.

Answer:

6.5 m/s

Explanation:

We are told that she is walking at 1.4 m/s and accelerates at 0.20 m/s².

Thus;

Initial velocity; u = 1.4 m/s

Acceleration; a = 0.2 m/s²

Distance; s = 100 m

From Newton's equation of motion, we know that;

v² = u² + 2as

Where v is final velocity.

Thus;

v² = 1.4² + 2(0.2 × 100)

v² = 41.96

v = √41.96

v ≈ 6.5 m/s

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