What are the horizontal and vertical components of a cat's displacement when the cat has climbed 5 m
directly up a tree © = 90°? Sin90=1
cos90=0

Answers

Answer 1

The horizontal component of the cat's displacement, Dx = 0 m

The vertical component of the cat's displacement, Dy = 12 m

What are the horizontal and vertical components of a vector?

The horizontal component of a vector is the value of that vector acting in the horizontal direction.

The horizontal component of a vector is given as Dx.

Dx = Dcosθ

where;

D is the resultant of the vector

θ is the angle of the vector from the horizontal

The horizontal component of the cat's displacement is:

Dx = 12 * 0

Dx = 0 m

The vertical component of a vector is the value of that vector acting in the vertical direction.

The vertical component of a vector is given as Dy.

Dy = Dsinθ

where;

D is the resultant of the vector

θ is the angle of the vector from the horizontal

The vertical component of the cat's displacement is:

Dy = 12 * 1

Dy = 12 m

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Related Questions

Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which radiates with a peak wavelength of about 970 nm? K (b) Rigel, a bluish-white star in Orion, radiates with a peak wavelength of 145 nm. Find the temperature of Rigel's surface. K

Answers

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}[/tex]

where,

T = Radiated surface temperature

[tex]\lambda_{max}[/tex] = peak wavelength

(a)

here,

[tex]\lambda_{max}[/tex] = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}[/tex]

T = 2987.6 k

(b)

here,

[tex]\lambda_{max}[/tex] = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

[tex]T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}[/tex]

T = 19986.2 k

just give me the right answer!
If the distance between two objects decreased, what would happen to the force of gravity between them?

It would increase.
It would stay the same.
It would depend on the speed.
It would decrease.

Answers

It would increase................

A bullet traveling at 5.0 x10^2 meters per is brought to rest by an impulse of 50 Newton*seconds. Find the mass of the bullet.

Answers

The bullet stops moving on hitting on a surface. Hence, the impulse here is equal to the momentum. Therefore, the mass of the bullet is 0.1 Kg.

What is impulse?

Impulse in physics is the change in momentum. It is the product of the force and change in time.

hence, impulse = f. dt

When the bullet is travelling with  a velocity of 500 m/s it has a momentum. When it brought to rest, momentum become zero. Thus, the momentum is equal to the impulse here.

Therefore, f.dt = m. v

f.dt = 50 N s

v = 500 m/s

m = 50 N s/500 m/s = 0.1 Kg

Therefore, the mass of the bullet is 0.1 Kg.

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HELPPP PLEASEEEEE, BRIANLEST WILL BE GIVEN ON CORRECT!​

Answers

Answer:

a. 6000J or 6KJ

b. Force =600

A supercluster is 100 million light-years across. How long would it take light to travel from one edge of the supercluster to the center of the supercluster?

Answers

Answer:

50 million years

Explanation:

light years is the distance light travels in one year given that the supercluster is 100 million light years across the the distance to the center will be half that amount therefore the answer is 50 million years

Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is sitting three meters away from the fulcrum at the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. The mass of the board is evenly distributed so that its center of mass is over the fulcrum. What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground? 
a. 4m
b. 5m
c. 2m
d. 3m​

Answers

Answer:

Option A. 4 m

Explanation:

Please see attached photo for diagram.

In the attached photo, X is the distance from the centre to which the student on the right must sit in order to balance the seesaw.

Clockwise moment = X × 45

Anticlock wise moment = 3 × 60

Clockwise moment = Anticlock wise moment

X × 45 = 3 × 60

X × 45 = 180

Divide both side by 45

X = 180 / 45

X = 4 m

Thus, the student must sit at 4 m from the centre.

The distance from the center the student on the right will be if they want the

seesaw to stay parallel to the ground is 4m

The question tells us that X is the distance to the center , each side of the

see saw is 5m with total length being 10m. This is explained in the

attached picture.

Clockwise moment = X × 45

Anticlock wise moment = 3 × 60

Clockwise moment = Anticlock wise moment

X × 45 = 3 × 60

X × 45 = 180

           = 180 / 45

            = 4m

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HELPPP PLEASEEEEE, BRIANLEST WILL BE GIVEN ON CORRECT!​

Answers

Answer:

a. 25000 J

b. 2500 J/s

Explanation:

Given,

Distance ( s ) = 50 m

Force ( f ) = 500 N

a.

To find : -

Work done ( W ) = ?

Formula : -

W = fs

W

= 500 x 50

= 25000 J

Therefore,

the work done by the force the horse exerts is

25000 J.

b.

To find : -

Power ( P ) = ?

Formula : -

W = Pt

P = W / t

P

= 25000 / 10

= 2500 J/s

Therefore,

the power produced if the movement took 10 s

is 2500 J/s.

Answer:

2500j/s

Explanation:

Kind of energy a piece of radioactive metal contains

Answers

Answer:

Radioactive materials give off a form of energy called ionizing radiation.

Electromagnetic waves are commonly referred to as _________
O electricity
O magnetism
O light

Answers

It’s going to be both answer A and B but if you can only answer one then it’s going to be B

Please help me thank you

Answers

The gravitational potential energy of a body is determined by the equation U=mgh, where m is the body's mass, g is its gravitational force, and h is its height. The relationship between mechanical energy and gravitational potential energy is described by the equation E=U+K.

What is the kinetic energy equation?

KE= 1/2 m v2

The relationship between kinetic energy and an object's mass and square of its velocity is direct: K.E. = 1/2 m v2.

We compute kinetic energy because...

It provides information on how a substance's mass impacts its velocities. Take this as a case study. A lorry and a sleek vehicle powered by the same engine cannot go at the same pace because to their different designs.

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A 25 kg box of roofing shingles is lifted 15 meters vertically to the top of a roof. How much work is done on the box?

Answers

Important Formulas:

[tex]F=ma[/tex]

force(measured in newtons) = mass(measured in kg) * acceleration(measured in m/s^2)

[tex]w=Fd[/tex]

work(measured in joules) = force(measured in newtons) * distance(measured in meters)

__________________________________________________________

Given:

[tex]m=25kg[/tex]

[tex]d=15m[/tex]

[tex]w=?[/tex]

We know that the acceleration due to gravity is 9.8 m/s^2.

__________________________________________________________

Finding force:

[tex]F=ma[/tex]

[tex]F=25\times9.8[/tex]

[tex]F=245N[/tex]

__________________________________________________________

Finding work:

[tex]w=Fd[/tex]

[tex]w=245\times15[/tex]

__________________________________________________________

[tex]\fbox{w = 3675 Joules}[/tex]

Some bat species have auditory systems that work best over a narrow range of frequencies. To account for this, the bats adjust the sound frequencies they emit so that the returning, Doppler-shifted sound pulse is in the correct frequency range. As a bat increases its forward speed, should it increase or decrease the frequency of the emitted pulses to compensate?

Answers

Answer:

As a bat increases its forward speed, it should decrease the frequency of the emitted pulses to compensate.

decrease

Explanation:

Decreasing the frequency of the emitted pulse will help the bat reduce its frequency caused by its forward motion.  The forward motion shifts the bat's auditory frequency to a higher frequency; consequently, the bat should adjust downwards the frequency of the emitted pulse so the reflected pulse will be in the correct frequency range.

A 6 kg bowling ball is lifted 1.2 m into a storage rack. The acceleration of gravity is 9.8 m/s². Calculate the increase in the ball's potential energy. Answer in units of J.​

Answers

Answer:

70.56 J

Explanation:

The increase in potential energy of the ball can be calculated using the formula:

PE = mgh

where:

PE is the increase in potential energy

m is the mass of the ball (6 kg)

g is the acceleration of gravity (9.8 m/s²)

h is the height the ball is lifted (1.2 m)

Substituting in the values, we get:

PE = (6 kg)(9.8 m/s²)(1.2 m)

This simplifies to:

PE = 70.56 J

So the increase in the ball's potential energy is 70.56 J.

What are the differences between thermionic and photoelectric emission?​

Answers

Answer:

During thermionic emission, electrons are emitted from metal surfaces by providing heat energy, while during photoelectric emission, light energy is emitted when electrons are emitted from the surface of metal

Hope it helps :)

three way to calculate average are

Answers

mean, median, mode.

mean is most commonly used.
to get the mean value, you add up all the values and divide this total by the number of values.

example: what is the average of the values 8, 5, and 6

8+5+6 = 19

19/3 = 6.3

6.3 is your average

Why is Energy, Work and Power all Scalar Quantity?​

Answers

Answer:

Explanation:

We already know that Force is a vector. Weight being a force, is also a vector quantity. Displacement is distance in a specific direction, hence it is a vector quantity too. ... since energy, work done and time are scalar, Power is a scalar quantity

Wow we both legit have the same question thx for asking it

3 ) find the electrical force between the two charges Q1=3mc ,Q2=-6mc when they are 0.3 m parted ?
find the amount of the force when Q1 is doubled ?​

Answers

Answer:

  F = 3.6 10⁶ N

Explanation:

The expression for the electric force is

         F =  [tex]k \ \frac{q_1 q_2}{r^2}[/tex]

in this case it indicates that the charge q1 is doubled

       q₁ = 2   3 10⁻³ C

       q₁ = 6 10⁻³ C

   

let's reduce the magnitudes to the SI system

        q₂ = 6 10⁻³ C

        r = 0.3 m

let's calculate

       F = 9 10⁹ 6 10⁻³ 6 10⁻³ / 0.3²

       F = 3.6 10⁶ N

QUESTION 4
A student lifts a 400 N sandbag 2 meters off the ground. How much work, in joules, did the student perform?

Answers

Answer:

800J

Explanation:

W = Fs, Work equals force times displacement

in this case, the force is 400N and the displacement is 2 meters.

The regular SI unit for work is joules

A satellite of mass m is orbiting Earth in a stable circular orbit of radius R. The mass and radius of Earth are ME and RE , respectively. Express your answers to parts (a), (b), and (c) the following in terms of m, R, ME , RE , and physical constants, as appropriate.

a. Derive an expression for the speed of the satellite in its orbit.
b. Derive an expression for the total mechanical energy of the satellite-Earth system in its orbit.
c. Derive an expression for the period of the satellite’s orbit.

Answers

Answer:

a)   v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex],   b)  Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex], c)  T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]

Explanation:

a) For this exercise we must use Newton's second law with the gravitational force

          F = ma

          [tex]G \frac{m M_e}{(R+R_e)^2 }[/tex] = m a

the acceleration of the satellite is centripetal

          a = [tex]\frac{v^2}{(R+R_e)}[/tex]

we substitute

            [tex]G \frac{m M_e}{(R+R_e)^2 } = m \frac{v^2}{ (R+R_e)}[/tex]

          [tex]G \frac{M_e}{(R+R_e)}[/tex]  = v²

          v = [tex]\sqrt{G \frac{M_e}{(R+R_e)} }[/tex]

the distance is from the center of the earth

b) mechanical energy is the sum of kinetic energy plus potential energy

         Em = K + U

         Em = ½ m v² - G m M / (R + R_e)

we substitute the expression for the velocity

         Em = ½ m  [tex]G \frac{M_e}{(R+R_e)}[/tex]  - [tex]G \frac{M_e}{(R+R_e)}[/tex]  

         Em = - ½ m [tex]G \frac{M_e}{(R+R_e)}[/tex]  

c) as the orbit is circulating, the velocity modulus is constant

         v = d / t

in a complete orbit the distance traveled of the circle is

        d = 2π (R + R_e)

where time is called period

         v = 2π (R + R_e)

         T = 2π (R + R_e) / v

we substitute the speed value

        T = 2π (R + R_e) [tex]\sqrt{\frac{(R+R_e) }{G M_e } }[/tex]

        T = 2π [tex]\sqrt{\frac{ (R+R_e)^3 }{G M_e } }[/tex]

(a) An expression for the speed of the satellite in its orbit.

[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]

(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.

[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]

(c) An expression for the period of the satellite’s orbit.

[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]

What are satellites?

A satellite is a moon, planet or machine that orbits a planet or star. For example, Earth is a satellite because it orbits the sun

a) For this exercise we must use Newton's second law with the gravitational force

F = ma

[tex]ma =G\sqrt{\dfrac{mM_e}{(R+R_e)}[/tex]

the acceleration of the satellite is centripetal

[tex]a=\dfrac{v^2}{R+R_e}[/tex]

we substitute

[tex]G\dfrac{mM_e}{(R+R_e)}=m\dfrac{v^2}{(R+R_e)}[/tex]

[tex]G\dfrac{M_e}{(R+R_e)}=v^2[/tex]

[tex]v=\sqrt{G\dfrac{M_e}{(R+R_e)}[/tex]

b) mechanical energy is the sum of kinetic energy plus potential energy

        Em = K + U

[tex]Em =\dfrac{1}{2}m v^2 - \dfrac{G m M} {(R + R_e)}[/tex]

we substitute the expression for the velocity

[tex]E_m=\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}-G\dfrac{M_e}{(R+R_e)}[/tex]

[tex]E_m=-\dfrac{1}{2}G\dfrac{M_e}{(R+R_e)}[/tex]

c) as the orbit is circulating, the velocity modulus is constant

[tex]v=\dfrac{d}{t}[/tex]  

in a complete orbit the distance traveled of the circle is

[tex]d = 2\pi (R + R_e)[/tex]

where time is called period

[tex]v = 2\pi (R + R_e)[/tex]

[tex]T = \dfrac{2\pi (R + R_e)} { v}[/tex]

we substitute the speed value

[tex]T = 2\pi (R + R_e) . \sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]

[tex]T=2\pi\sqrt{\dfrac{(R+R_e)}{GM_e}[/tex]

(a) An expression for the speed of the satellite in its orbit.

[tex]V=\sqrt{G\dfrac{M_e}{R+R_e}[/tex]

(b) An expression for the total mechanical energy of the satellite-Earth system in its orbit.

[tex]E_m =\dfrac{1}{2}mG\dfrac{M_e}{(R+R_e)}[/tex]

(c) An expression for the period of the satellite’s orbit.

[tex]T=2\pi\sqrt\dfrac{(R+R_e)^3}{GM_e}[/tex]

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a defender running away from a goalkeeper at 5m/s is hit in the back by the goal kick. the ball stops dead and the players speed increases to 5.5m/s if the ball had a mass of 500g and the player mass is 70k how fast was the ball moving?

Answers

The goal kick strikes a defender racing at 5 m/s away from a goalkeeper in the back. The ball was moving at a speed of 70 m/s if it weighed 500g and the player's mass was 70k.

Why does the goalie in a football game pull his hands back after holding the ball that has been shot towards the goal?

The goalkeeper extends the amount of time he has to hold the ball by pulling his hands back. He lessens the force (rate of change of momentum) the football exerts on him by lengthening the time.

Write the formula for momentum and define it.

A vector quantity, momentum. A body's momentum is equal to P=m.v if it is travelling at a speed of v while having mass m. P=mv describes the magnitude of the momentum. Because momentum is a function of mass and velocity, its unit is the product of those two quantities.

By using conservation of momentum,

Initial momentum = Final momentum

(5 x 70) + (v x 0.5) = (5.5 x 70) + (0 x 0.5)

350 + 0.5v = 385 + 0

0.5v = 35

v = 70 m/s

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You are riding your bicycle down the street at a speed of 14 m/s. Your bicycle frame has a mass of 6.6 kg, and each of its two wheels has mass 2.2 kg and radius 0.35 m. Each wheel can be thought of as a hollow hoop (assuming that the rim has much larger mass than the spokes). What is the total kinetic energy of the bicycle (in Joules), taking into account both the translational and rotational motion

Answers

Answer:

1078 Joules

Explanation:

The computation of the total kinetic energy of the bicycle is shown below:

Given that

mass of bicycle's frame (m) = 6.6 kg

mass of each wheel (M) = 2.2 kg

radius of each wheel (r) = 0.35 m

And, the linear speed (v) = 14 ms

Now

As we know that

Angular velocity (ω) = v ÷ r

= 140 ÷ .35

= 40 rads

Total kinetic energy = translational kinetic energy + rotational kinetic energy

= (1 ÷2 × m  × v^2) + (2 × 1 ÷ 2×I × ω^2)

= (0.5 × 6.6 × [14]^2) + (M × r2 × ω^2)

= 646.8 + (2.2 × 0.35 × 0.35 × [40]^2)

= 646.8 + 431.2

= 1078 Joules

what energy transformation is preformed by a radio

Answers

Answer:

Chemical energy to sound energy to heat energy

Please give me the answer to this question

Answers

There is not enough information to determine the work done. Option iv

What is the work done?

Let us note that we say that there is work done when the force that has been applied moves a distance in the direction of the force. In this case, we have been told that there is the combination of the works that is done by the object.

Now, we also have to note that we do not have other information to determine the work done such as the magnetic feild and the mass of the electron. All these are lacking in the question.

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A boat has a speed of 20m/s in still waters. what is the speed in a river flowing due east with a velocity of 5m/s if the boat is heading north. How far from a point directly north on the other side of the river will the boat arrive if the width of the river is 10m.​

Answers

The boat will arrive 2.5 meter in east far from a point directly north on the other side of the river.

What is velocity?

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction.

A boat has a speed of 20m/s in still waters.

The velocity of river is 5m/s in east.

The width of the river is = 10 meter.

Hence, time taken by the board to cross the river = 10/20 second

= 0.5 second.

In this time, the board will move in east direction   = 5×0.5 meter = 2.5 meter.

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An 80-kg man balances the boy on a teeter-totter as shown. What is the approximate mass of the boy? What, approximately, is the magnitude of the downward force exerted on the fulcrum? Ignore the weight of the board.
I know the first answer is 20 kg and the second answer is 1000N. Have no idea how to get these answers though. Please help? Will rate!!

Answers

The boy weighs at about 80 kg. When, as illustrated, a N 784 guy balances the youngster on a teeter-totter.

What is ideal interpretation of mass?

A way to determine how much material, in mass, makes up a physical body.

In classical mechanics, an object's mass is essential to Newton's equations of motion because it affects the force required to accelerate it and, subsequently, how much inertia it has.

Language expert Peter Trudgill claims that the name "Tittermatorter" from the Norfolk language is its origin. Trudgill can trace all of his great-great-grandparents to a tiny region of eastern Norfolk.

As per the question:

Weight is come under gravity and become force

80g = mg

m = 80

Force = 80(9.8) = 784N

Thus, the answer is 784N.

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1. Three confused sled dogs are trying to pull a sled across the Alaskan snow. Tim pulls east with a force of 42 N; Sam also pulls east, but with a force of 53 N; and big Ethan pulls west with a force of 67 N. What is the net force on the sled? then find the acceleration?

2. A 944-kg dragster, starting from rest, attains a speed of 31,2 m/s in 0.670 s. a. Find the average acceleration of the dragster.

b. What is the magnitude of the average net force on the dragster during this time?

c. What horizontal force does the seat exert on the driver if the driver has a mass of 68.0 kg?

Please try to atleast answer one

Answers

Answer:

Explanation:

According to the statement, three confused sleigh dogs are trying to pull a sled across the Alaskan snow.  

Forces in same direction gets added , so 35N + 42N=77N  and the Net Force is 77N -53N as it is acting in opposite direction.

Net force is 25N in east to the maximum without any hassle.

-------------------------------------------------------------------------------------------------

thankyou : )

with what speed will a clock have to be moving in order to run at a rest that is one half the rate of clock at rest

Answers

The speed at which a clock would have to be moving in order to run at half the rate of a clock at rest depends on the theory of relativity that you are using.

What is the clock  speed about?

In special relativity, time dilation is the phenomenon where time appears to pass differently for objects in motion relative to an observer at rest.

According to the theory, time appears to slow down for an object as it approaches the speed of light. The rate at which time appears to pass for an object is given by the equation:

T' = T / [tex]\sqrt{(1 - (v^2 / c^2))}[/tex]

Where T is the time as measured by an observer at rest, T' is the time as measured by an observer moving relative to the object, v is the velocity of the object, and c is the speed of light.

In addition, this is a theoretical scenario, practically in order to measure time dilation in a laboratory, it is required a very high precision of measurements, that are currently not possible.

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What are the magnitude and direction of the current in the 20 Ω resistor in (Figure 1)?
Express your answer with the appropriate units. Enter positive value if the current is clockwise and negative value if the current is counterclockwise.

Answers

To solve this we must be knowing each and every concept related to  net current. Therefore, the circuit's overall net current is 0.1 A, flowing clockwise.

What is the net current?

The term "net current" refers to the total currents after taking current direction into account. If a node sends 1 mA inside and 5 mA out, the total current equals 4 mA out.

The sum of all entering currents must match the sum of all exiting currents, according to KCL. According to Ohm's Law, the current flowing is equivalent to the combined voltage dividing by the entire resistance. In contrast to a series circuit, where the total current is the same across all of the components, the total current inside a parallel connection is really only equal to the particular current in that part of the circuit.

Net current

i = V / Ri

 = (9 - 6) / (10 + 20)i

 = 0.1 A

in clockwise way

Therefore, the circuit's overall net current is 0.1 A, flowing clockwise.

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What is the resulting acceleration when a net force of 18 N is applied to a 3.0 kg mass?
O A. 6.0 m/s2
C. 54 m/s2
O B. 15 m/s2
O D. 9.8 m/s2

Answers

Answer:

F =ma

a=f/m =18/3= 6m/m²

Explanation:

the resulting acceleration will 6 m/s² because the question asking us to solve for the acceleration.

In each of the four situations below an object is experiencing (nearly) uniform circular motion. State what force is providing the centripetal force required to keep the object moving in a circle: a. A car driving around a track. b. A ball being swung on the end of a string. c. The moon orbiting the earth. d. A rotating wheel.

Answers

Answer:

Explanation:

Given that a centripetal force is a form of force that gives rise or causes a body to move in a curved path.

Hence;

1. When a car is being driven around a track, it is the FORCE OF FRICTION that is acting upon the turned wheels of the vehicle, which transforms into the centripetal force required for circular motion.

2. When a ball being is swung on the end of a string, TENSION FORCE acts upon the ball, which transforms the centripetal force required for circular motion.

3. When the moon is orbiting the earth, it is the FORCE OF GRAVITY acting upon the moon, which transforms the centripetal force required for circular motion.

4. A rotating wheel on the other hand has NO centripetal force because centripetal force is pull towards the center of a motion. However the speed of the object is tangent to the circle, while the direction of the force is also perpendicular to the direction of the rotating wheel.

The force that provides the centripetal force in each of the given situations are;

A) Friction Force

A) Friction ForceB) Tension Force

A) Friction ForceB) Tension ForceC) Force of gravity

A) Friction ForceB) Tension ForceC) Force of gravity D) No centripetal force

When an object is in circular motion, the force that keeps it moving round the circle while centrifugal force is the one that tries to pull the object away from the center.

A) When a car is driving around a track, there is a frictional force between the tires of the car and the track that acts on the vehicle to keep it in that circular motion. This frictional force is the centripetal force required to keep the vehicle in circular motion.

B) When a ball is swing on the end of a string, there is an upward force called tension force that acts on the ball to keep it swinging in circular motion. Thus, the centripetal force here is provided by the tension force.

C) When the moon is orbiting the earth, there is a force of gravity exerted by the earth on the moon that keeps the moon in a circular motion about Earth instead of moving in a straight line.

D) For a rotating wheel, the centripetal force does not do any work. The reason for that is because the centripetal force points toward the center of the circle, and as a result it means that the velocity of the rotating wheel is tangent to the circle.

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