In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point 2.5 m from his foot at the edge of the pool.
Where does the spot of light hit the bottom of the pool, measured from the bottom of the wall beneath his foot, if the pool is 2.1 m deep pool?

Answers

Answer 1

Answer:

4.4 m

Explanation:

We are told the light from his flashlight, 1.3 m above the water level. Thus; h1 = 1.3m

Also,we are told that the light shone 2.5 m from his foot at the edge of the pool. Thus, L1 = 2.5 m

Angle of incidence θ1 is given by;

tan θ1 = L1/h1

tan θ1 = 2.5/1.3

tan θ1 = 1.9231

θ1 = tan^(-1) 1.9231

θ1 = 62.53°

Using Snell's law, we can find the angle of refraction from;

Sin θ2 = (η_air/η_water) Sin θ1

Where;

η_air is Refractive index of air = 1

η_water is Refractive index of water = 1.33

Thus;

Sin θ2 = (1/1.33) × sin 62.53°

Sin θ2 = 0.6671

θ2 = sin^(-1) 0.6671

θ2 = 41.84°

We want to find where the spot of light hit the bottom of the pool if the pool is 2.1 m deep. Thus, h2 = 2.1 m

Now, the spot can be found from;

L = L1 + L2

Where L2 = (h2) tan θ2

L = 2.5 + 2.1 tan 48.84

L = 2.5 + (2.1 × 0.8954)

L ≈ 4.4 m


Related Questions

Which of the following is generally true for people in their old age?

Answers

Answer:As people get older, most become better able to regulate negative feelings and emphasize the positive.

Explanation: With age, your skin thins and becomes less elastic and more fragile, and fatty tissue just below the skin decreases. You might notice that you bruise more easily. Decreased production of natural oils might make your skin drier. Wrinkles, age spots and small growths called skin tags are more common.

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.80×106 N , one at an angle 14.0 ∘ west of north, and the other at an angle 14.0 ∘ east of north, as they pull the tanker a distance 0.800 km toward the north. Part A What is the total work done by the two tugboats on the supertanker?

Answers

Answer:

W = 1,049 10⁹ J

Explanation:

Work is defined by the relation

         W = F. d = F d cos θ

where tea is the angle between the forces and the displacement.

The total work is the sum of the work of each tug.

Tug 1

       W₁ = F d cos θ₁

 

the angle measured from the positive side of the x-axis is

       θ₁ = 14 + 90 = 104º

           

tugboat 2

             W₂ = F d cos θ₂

             θ₂ = 14

we substitute

             W = F d cos θ₁ + F d cos θ₂

             W = F d (cos θ₁ + cos θ₂)

               

let's calculate

             W = 1.80 10⁶  800 (cos 104 + cos 14)

             W = 1,049 10⁹ J

An object is accelerated by a net force in which direction?
A. at an angle to the force
B. in the direction of the force
C. in the direction opposite to the force
D. Any of these is possible.

Answers

Answer:

B. in the direction of the force

Explanation:

Sana nakatulong

On a cold winter day, a steel metal fence post feels colder than a wooden fence post of identical size because: a. The specific heat capacity of steel is higher than the specific heat capacity of wood. b. The specific heat capacity of steel is lower than the specific heat capacity of wood. c. Steel has the ability to resist a temperature change better than wood. d. The mass of steel is less than wood so it loses heat faster. Selected:e. Two of the above statements are true.

Answers

Answer:

The specific heat capacity of steel is lower than the specific heat capacity of wood

Explanation:

THERE IS ONLY 1 ON MY assignment i geot dis right  please brainlyist

The specific heat capacity of steel is lower than the specific heat of a piece of wood. Therefore, option (2) is correct.

What is the specific heat capacity?

Specific heat can be defined as the heat energy required to change the temperature of one unit mass of a substance of a constant volume by 1 °C. The S.I. unit of the specific heat capacity of a material is KJ/Kg.

The thermal capacity of a material is defined as a physical property of a substance. The amount of heat is given to a given mass to create a change in unit temperature.

The mathematical expression of specific heat capacity can be written as :

Q = m C ΔT      Where C is the heat capacity.

The specific heat capacity is an intensive property of a substance as it does not depend upon the size of the material.

A steel metal fence post feels colder than a wooden fence post of similar size because the specific heat of steel is lower than the specific heat capacity of wood.

Learn more about specific heat, here:

brainly.com/question/11297584

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g 4.86 Separators are used to separate liquids of diff erent densities, such as cream from skim milk, by rotating the mixture at high speeds. In a cream separator, the skim milk goes to the outside while the cream migrates toward the middle. A factor of merit for the centrifuge is the centrifugal acceleration force (RCF), which is the radial acceleration divided by the acceleration due to gravity. A cream separator can operate at 9000 rpm (rev/min). If the bowl of the separator is 20 cm in diameter, what is the centripetal acceleration if the liquid rotates as a solid body, and what is the RCF

Answers

Answer:

Centripetal Acceleration = 88826.44 m/s²

RCF = 9054.7

Explanation:

First, we will find the value of the centripetal acceleration by using the following formula:

[tex]Centripetal\ Acceleration = \frac{v^2}{r}\\[/tex]

where,

v = linear speed of liquid or separator = rω

ω = angular speed of liquid or separator = (9000 rpm)(2π rad/rev)(1 min/60 s) = 942.48 rad/s

r = radius of seperator = diameter/2 = 20 cm/2 = 10 cm = 0.1 m

Therefore,

[tex]Centripetal\ Acceleration = \frac{(r\omega)^2}{r}\\Centripetal\ Acceleration = r\omega^2\\Centripetal\ Acceleration = (0.1\ m)(942.48\ rad/s)^2\\[/tex]

Centripetal Acceleration = 88826.44 m/s²

Now, for the RCF:

[tex]RCF = \frac{Centripetal\ Acceleration}{g}\\RCF = \frac{88826.44\ m/s^2}{9.81\ m/s^2}\\[/tex]

RCF = 9054.7

A 430 kg motorcycle starts from rest and accelerates to a speed of 12 m/s.
Calculate the net work done on the motorcycle.
a. 42 kJ
b. 31 kJ
c. 38 kJ
d. 35 kJ
e. none of these

Answers

Answer:

Vi = 0

Vf = 12 m/s

ΔV = Vf - Vi

ΔV = 12 m/s

Change in kinetic energy

ΔKE = Kf - Ki

ΔKE = 1/2 mv^2 - 0

ΔKE = 1/2 * 430 kg * (12 m/s)^2

ΔKE = 30,960

ΔKE = 30,960 joules

 

Work = ΔK

Work = 30,960 J

A group of particles of total mass 48 kg has a total kinetic energy of 320 J. The kinetic energy relative to the center of mass is 80 J. What is the speed of the center of mass?

Answers

K=.5mv^2 + relative Ke
320=.5(48)v^2+80
V=3.16

Find the momentum of a 15 kg object traveling at 7 m/s

What is the momentum
What is the velocity
What is the mass

What equation did you use to solve?

Answers

Find the momentum of a 15 kg object traveling at 7 m/s.

The momentum of an object is found by using the following formula:

[tex]\displaystyle p=mv[/tex]

P is the momentum and is measured in kg · m/sm is the mass and is measured in kgv is the velocity and is measured in m/s

In this question, the object is 15 kg and is travelling at 7 m/s. That means the mass is 15 kg and the velocity is 7 m/s.

Since all the needed variables are found, substitute it into the equation:

[tex]\displaystyle p=mv \rightarrow p=15 \times 7[/tex]

Multiply:

[tex]\displaystyle p=105\ kg \times m/s[/tex]

__________________________________________________________

What is the momentum? 105 kg · m/s

What is the velocity? 7 m/s

What is the mass? 15 kg

What equation did you use to solve? p = mv

__________________________________________________________

Two blocks (with masses of 2.0 kg and 4.0 kg) are on a bench tied together with string. They are being pulled to the right with a force of 30N. They each experience a friction force between the block and the bench.
(Refer to image)

The 2 kg block experiences a friction force with a friction coefficient of 0.30 and the 4 kg experiences a friction with a friction coefficient of 0.20.

Assume that g (the acceleration due to gravity) is 10.0 m/s/s.

Find the magnitude of the friction forces. Find the magnitude of the acceleration of the blocks. Use these answers to help you find the answer to the question:

What is tension in the string connecting the two blocks? (Submit just this answer in Newtons)

Answers

Answer:

T = 34/3 N

Explanation:

Magnitude of the friction force on 2kg block = 0.3x10x2 = 6N

Magnitude of the friction force on 4kg block = 0.2x10x4 = 8N

Magnitude of the acceleration of the blocks

F = ma

30 - 8 - 6 = (2+4)a

a = 8/3 m s^-2

Tension in the string connecting the two blocks

Consider the 2kg block,

T - f = ma

T - 6 = 2(8/3)

T = 34/3 N

Which of the following have frequencies greater than orange light Your answer:
radio waves
purple light
ultraviolet rays
red light
green light
gamma rays
microwaves
infrared rays

Answers

Answer:

Gamma Rays have the highest frequencies

Explanation:

This is because Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies compared to the other light frequency. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation which means the answer has to be gamma rays. Brainly Please!!!! Here are screenshots that may help

What is diffraction of light

Answers

Answer:

According to "http://ww2010.atmos.uiuc.edu"  Diffraction is the slight bending of light as it passes around the edge of an object.

Some examples of Light Defraction would be..

-CD reflecting rainbow colours

-Sun appears red during sunset

-From the shadow of an object

(will give brainliest to whoever is correct and shows reasoning) What is the acceleration of an object that has a velocity of 60m/s and is moving in a circle of radius 50m?

Answers

Answer:

5.0/s

Explanation:

Answer:

b and a it is this that abewsr

A light bulb is shown below, shining into a concave mirror, with its original light lines visible. Which statement best explains why the image of the bulb appears behind the mirror, as shown? .
A. The original light comes from there.
B. The reflected light comes from there.
C. The original light appears to come from there. This is often indicated with dotted apparent light lines.
D. The reflected light appears to come from there. This is often indicated with dotted apparent light lines.

Answers

Answer:

the answer is c

Explanation:

see the light appears from there and with the dotted lines you can clearly see the green line touches the dot okay, then the light appears smaller because of water and light source

The optics of your visual system have a total refractive power of about +60 D—about +20 D from the lens in your eye and +40 D from the curved shape of your cornea. Surgical procedures to correct vision generally do not work on the lens; they work to reshape the cornea. In the most common procedure, a laser is used to remove tissue from the center of the cornea, reducing its curvature. This change in shape can correct certain kinds of vision problems.
The length of your eye decreases slightly as you age, making the lens a bit closer to the retina. Suppose a man had his vision surgically corrected at age 30. At age 70, once his eyes had decreased slightly in length, he would be:________.
A. Nearsighted.
B. Farsighted.
C. Neither nearsighted nor farsighted.

Answers

Answer:

Farsighted

Explanation:

Farsightedness also known as hypermetropia is caused by the eye being too short(the eye shortens with advancing age) or the crystalline lines not being sufficiently convergent.

A farsighted person can see far objects but can not see nearby objects. His near point is now farther than the 25 cm near point of a normal eye. Images are formed some distance behind the retina.

This eye defect is corrected by the use of a converging lens to reduce the divergence of the rays entering the eye from an object.

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 38.0 ∘ above the horizontal. The glider has mass 9.00×10−2 kg. The spring has 590 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.70 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring.

Required:
a. What distance was the spring originally compressed?
b. When the glider has traveled along the air track 0.80 m from its initial position against the compressed spring, is it still in contact with the spring? What is the kinetic energy of the glider at this point?

Answers

Answer:

x = 0.056 m

ΔKE = 0.489 J

Explanation:

Given that

Angle, θ = 38°

Length, L = 1.7 m

Mass, m = 0.09 kg

Spring constant, K = 590 N/m

If we use the Work-Energy theorem, then we know that Potential Energy, PE = Kinetic Energy, KE

This is mathematically written as

1/2kx² = mgH

The height, H we can get by using the relation

H = L.Sinθ

H = 1.7 * Sin 38

H = 1.7 * 0.6157

H = 1.047 m

Next, we use the Work-Energy theorem

1/2kx² = mgH

1/2 * 590 * x² = 0.09 * 9.8 * 1.047

295 * x² = 0.9234

x² = 0.9235 / 295

x² = 0.00313

x = √0.00313

x = 0.056 m

If the spring loses contact at x = 0.056, definitely, it will also lose contact at x = 0.8

Then we use the formula

ΔKE = mg(H - H1)

ΔKE = mg(xsinθ - x2.sinθ)

Where, x = 1.7 , x2 = 0.8

ΔKE = 0.09 * 9.8 (1.7 * sin 38 - 0.8 * sin 38)

ΔKE = 0.882(1.047 - 0.493)

ΔKE = 0.882 * 0.554

ΔKE = 0.489 J

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine:

Answers

This question is incomplete, the complete question is;

Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).

Answer:

the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

Explanation:

Given the data in the question and image below and as illustrated in the second image;

distance S = 40 m

V[tex]_B[/tex] = 54 km/hr

V[tex]_A[/tex] = 72 km/hr

α = 100 m

now, angular velocity of Bxy will be;

ω[tex]_B[/tex] = V[tex]_B[/tex] / α

so, we substitute

ω[tex]_B[/tex] = ( 54 × 1000/3600) / 100

ω[tex]_B[/tex] = 15 / 100

ω[tex]_B[/tex] = 0.15 rad/s

Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s

What is the weight of a 48kg rock?

Answers

Answer:

48kg

Explanation:

A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4
kg, collides with a smaller box with a mass of 2 kg that is stationary. The boxes stick
together. What is the velocity of the two combined masses after collision?
8 m/s
O m/s
12 m/s
4 m/s
us 12:18

Answers

Answer= 8m/s

Because total Momentum before= total momentum after

Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48

Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u


Mb=Ma
48=6u
u=8m/s

Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.6-V battery, the current from the battery is 2.07 A. When the resistors are connected in parallel to the battery, the total current from the battery is 8.98 A. Determine R1 and R2. (Enter your answers from smallest to largest.)

Answers

Answer:

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193

Explanation:

We are told that when R1 and R2 are connected in series, the voltage is 12.6 V and the current is 2.07 A.

Formula for resistance is;

R = V/I

R = 12.6/2.07

R = 6.087 ohms

Since R1 and R2 are connected in series.

Thus; R1 + R2 = 6.087 ohms

R1 = 6.087 - R2

We are also told that when they are connected in parallel, the current is 8.98 A.

Thus, R = 12/8.98

R = 1.403 ohms

Thus;

(1/R1) + (1/R2) = 1/1.403

Let's put 6.087 - R2 for R1;

(1/(6.087 - R2)) + (1/R2) = 1/1.403

Multiply through by 1.403R2(6.087 - R2) to get;

1.403R2 + 1.403(6.087 - R2) = R2(6.087 - R2)

Expanding gives;

1.403R2 + 8.54 - 1.403R2 = 6.087R2 - (R2)²

(R2)² - 6.087R2 + 8.54 = 0

Using quadratic formula, we have;

R2 = 2.193 ohms or 3.894 ohms

Thus,

R1 = 6.087 - 2.193 or R1 = 6.087 - 3.894

R1 = 3.894 or 2.193

When R1 = 2.193, R2 = 3.894

When R1 = 3.894, R2 = 2.193

A planet of mass M has a moon of mass m in a circular orbit of radius R. An object is placed between the planet and the moon on the line joining the center of the planet to the center of the moon so that the net gravitational force on the object is zero. How far is the object placed from the center of the planet

Answers

Answer:

r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]

Explanation:

Let's apply the universal gravitation law to the body (c), we use the indications 1 for the planet and 2 for the moon

          ∑ F = 0

           -F_{1c} + F_{2c} = 0

             F_{1c} = F_{2c}

let's write the force equations

             [tex]G \frac{m_c M}{r^2} = G \frac{m_c m}{(d-r)^2}[/tex]

where d is the distance between the planet and the moon.

              [tex]\frac{M}{r^2} = \frac{m}{(d-r)^2}[/tex]

             (d-r)² = [tex]\frac{m}{M} \ \ r^2[/tex]  

              d² - 2rd + r² = \frac{m}{M} \ \ r^2

              d² - 2rd + r² (1 - [tex]\frac{m}{M}[/tex]) = 0

              (1 - [tex]\frac{m}{M}[/tex])  r² - 2d r + d² = 0

we solve the second degree equation

              r = [2d ± [tex]\sqrt{ 4d^2 - 4 ( 1 - \frac{m}{M} ) }[/tex] ] / 2 (1- [tex]\frac{m}{M}[/tex])

              r = [2d ±  2d [tex]\sqrt{ \frac{m}{M} }[/tex]] / 2d (1- [tex]\frac{m}{M}[/tex])

              r =[tex]\frac{ 1 \pm \sqrt{ \frac{m}{M} } }{1 - \frac{m}{M} }[/tex]

there are two points for which the gravitational force is zero

The distance between object from planet will be "[tex]\frac{R}{[1+\sqrt{\frac{m}{M} } ]}[/tex]".

According to the question,

Let,

Object is "x" m from planet center = R - xGravitational force = 0Mass of object = m₁

As we know,

→ [tex]Prerequisites-Gravitational \ force = \frac{GMm}{r^2}[/tex]

Now,

→ [tex]\frac{GMm_1}{x^2} = \frac{Gmm_1}{(R-x)^2}[/tex]

→ [tex]\frac{(R-x)^2}{x^2} = \frac{m}{M}[/tex]

→     [tex]\frac{R-x}{x} =\sqrt{\frac{m}{M} }[/tex]

→          [tex]x = \frac{R}{[1+ \sqrt{\frac{m}{M} } ]}[/tex]

Thus the answer above is appropriate.          

Learn more:

https://brainly.com/question/11968775

Force exerted on a body changes it's

Answers

Momentum is the answer

If you push with a power of 20 Watts
on a 150 Newton object, how long would
it take to push it over the 4.3 m?

Answers

Answer:

32.25 s

Explanation:

From the question,

P = W/t.............. Equation 1

Where P = Power, W = work done, t = time.

But

W = F×d................. Equation 2

Where F = force and d = distance

Substitute equation 2 into equation 1

P = F×d/t............... Equation 3

make t the subject of euqation 3

t = (F×d)/P............. Equation 4

Givn: F = 150 N, d = 4.3 m, P = 20 watts.

Substitute these values into equation 4

t = (150×4.3)/20

t = 32.25 s

Solve the below problems being sure to provide the correct significant figures.

1) 1000 ÷ 4.886 = __________

2) 240 ÷ 12.3 = __________

3) 80 x 4.6 = __________

4) 4.527 x 30 = __________

5) 86 x 63.855 x 8000 = __________

6) 700 x 91.186 = __________

7) 7.1 x 348 = __________

8) 50 ÷ 29.1 = __________

9) 98.773 x 24.891 x 409 = __________

10) 0.065 x 3 x 3007 = __________

Answers

Answer:

1) 204.6663938

2) 19.51219512

3) 368

4) 135.81

5) 43932240

6) 63830.2

7) 2470.8

8) 1.718213058

9) 1005550.526

10) 586.365

Most of the questions you asked were in repeating decimal form.

Explanation:

Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?

Answers

Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.

The speed limmit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?

Answers

I uploaded the answer to a file hosting. Here's link:

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A 40 kg rock is rolling toward a town at 4 m/s after an earthquake. Calculate the KE.

Be sure to show your work and include units!

Use the formula KE = 1/2mv2 will brainlist

Answers

Answer:

320 J

Explanation:

From the question,

KE = 1/2mv².................. Equation 1

Where KE = Kinetic Energy, m = mass of the rock, v = velocity of the rock

Given: m = 40 kg, v = 4m/s

Substitute these values into equation 1

KE = 1/2(40)(4²)

KE = 20×16

KE = 320 J

Hence the kinetic energy of the rock is 320 J

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One of the greatest dangers in a tornado is from flying objects. A 15 pound piece of lumber can turn into a flying missile that could severely damage walls and homes. A piece of steel with a mass of 200 pounds and travelling at the same velocity would cause even more damage. Select any evidence from the list below that you could use to explain why a 200 pound piece of steel would cause more damage than a 15 pound piece of wood travelling at the same velocity.

As the kinetic energy of an object increases, the force it can exert on another object decreases.

As the kinetic energy of an object increases, the force it can exert on another object increases.

Objects with more mass have less kinetic energy.

Objects with more mass have more kinetic energy.

As the velocity of an object increases, its kinetic energy decreases.

As the velocity of an object increases, its kinetic energy increases.

Answers

I uploaded the answer to a file hosting. Here's link:

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Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength ?

Answers

Complete question:

Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength  at the midpoint between the two rings ?

Answer:

The electric field strength at the mid-point between the two rings is zero.

Explanation:

Given;

diameter of each ring, d = 10 cm = 0.1 m

distance between the rings, r = 21.0 cm = 0.21 m

charge of each ring, q = 40 nC = 40 x 10⁻⁹ C

let the midpoint between the two rings = x

The electric field strength  at the midpoint between the two rings is given as;

[tex]E_{mid} = E_{right} +E_{left}\\\\E_{right} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } \\\\E_{leftt} = -\ \frac{KQ}{(x^2 + r^2)^\frac{2}{3} }\\\\E_{mid} = \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } - \frac{KQ}{(x^2 + r^2)^\frac{2}{3} } = 0[/tex]

Therefore, the electric field strength at the mid-point between the two rings is zero.

A uniform-density 7 kg disk of radius 0.20 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 42 N through a distance of 0.9 m. Now what is the angular speed

Answers

Answer:

The angular speed is 23.24 rad/s.

Explanation:

Given;

mass of the disk, m = 7 kg

radius of the disk, r = 0.2 m

applied force, F = 42 N

distance moved by disk, d = 0.9 m

The torque experienced by the disk is calculated as follows;

τ = F x d = I x α

where;

I is the moment of inertia of the disk = ¹/₂mr²

α is the angular acceleration

F x r = ¹/₂mr² x α

The angular acceleration is calculated as;

[tex]\alpha = \frac{2Fr}{mr^2} \\\\\ \alpha = \frac{2F}{mr}\\\\\alpha = \frac{2 \times 42 }{7 \times 0.2} \\\\\alpha = 60 \ rad/s^2[/tex]

The angular speed is determined by applying the following kinematic equation;

[tex]\omega _f^2 = \omega_i ^2 + 2\alpha \theta[/tex]

initial angular speed, ωi = 0

angular distance, θ = d/r = 0.9/0.2 = 4.5 rad

[tex]\omega _f^2 = 2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta} \\\\\omega _f = \sqrt{2 \times 60 \times 4.5} \\\\\omega _f = 23.24 \ rad/s[/tex]

Therefore, the angular speed is 23.24 rad/s.

What would happen if the molecules in a sample moving entirely ?

Answers

Answer:

Molecular scale. The story begins a long time ago  

when the idea that molecules are in constant motion  

was first discovered. Part of the evidence that you can  

see in everyday life was discovered by Robert Brown  

about 150 years ago when he used a microscope to  

watch how tiny dust particles move.

So how fast do molecules move? It all depends upon  

the molecule and its state: molecules in a solid state  

move slower than in a liquid state, and much slower  

than gas molecules. One estimate puts gas molecules  

in the range of 1,100 mph at room temperature. Cool  

them down to almost absolute zero and they slow  

down to less than 0.1 mph (slower than the average  

couch potato). The fact that they are always moving  

makes it a challenge to see molecules and make stuff  

out of them, but it’s a challenge that scientists  

work hard to figure out.

Explanation:

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