A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.00 m above the ground. The string breaks and the ball lands 2.30 m (horizontally) away from the point on the ground directly beneath the ball's location when the string breaks. Find the centripetal acceleration in m/s2 of the ball during its circular motion.

Answers

Answer 1

Answer:

86.7 m/s²

Explanation:

Given that

Radius of the circle, r = 0.3 m

Height of the plane above the ground, = 1 m

Distance at which the ball lands, d = 2.3 m

To solve this question, we would use one of the principles of projectile motion.

v = ut + 1/2at²

-1 = 0 + 1/2 * -9.8 * t²

-1 = -4.9 t²

t² = 1/4.9

t² = 0.204

t = √0.204

t = 0.45 s

Then again, it's constant horizontal speed is

v(x) = x/t

v(x) = 2.3 / 0.45

v(x) = 5.1 m/s

And finally, the centripetal acceleration is gotten using the formula

a = v²/r

a = 5.1² / 0.3

a = 26.01 / 0.3

a = 86.7 m/s²


Related Questions

What x rays travel at the speed of

Answers

What is the speed? Like we need the pic or something

Missy Diwater, the former platform diver for the Ringling Brother's Circus had a kinetic energy of 15,000 J just prior to hitting the bucket of water. If Missy's mass is 50 kg, then what is her speed?

Answers

Answer:

24.5 m/s

Explanation:

KE=1/2mv^2

15000=1/2(50)v^2

30000=(50)v^2

600=v^2

sqrt600=v

v=24.5 m/s!!

What are characteristics of wide-angle lenses?

Answers

One of the characteristics of a wide-angle lens is the ability to capture the expanse of a scene in an image with its wide angle of view. Also, it is able to create an intriguing deformed effect when you photograph the subject from a close distance. These characteristics can be utilised to bring out the shape of a subject or the depth of a landscape. That being said, care is needed as distortion is more likely to occur toward the edges of the image. To make effective use of a wide-angle lens, it is important to control both the angle of view as well as distortions in the image at the same time. By doing so, you will be able to add a dynamic touch to your photographic expression.

An official major league baseball has a mass of 0.14 kg. A pitcher throws a 40 m/s fastball which is hit by the batter straight back up the middle at a speed of 46 m/s.
a) What is the change in momentum of the ball during the collision with the bat?
b) If this collision occurs during a time of 0.012 seconds, what is the average force exerted by the bat on the ball?

Answers

Answer:

(a) The change in momentum is 12.04 kg-m/s

(b) The force exerted by the bat is 1003.33 N

Explanation:

Given that,

The mass of a ball, m = 0.14 kg

Initial speed of the ball, u = 40 m/s

Final speed of the ball, v = -46 m/s

(a) The change in momentum of the ball during the collision with the bat is given by :

[tex]\Delta p=m(v-u)\\\\=0.14(-46-40)\\\\=-12.04\ kg-m/s[/tex]

(b) Time for collision, t = 0.012 s

Now the force can be calculated as follows :

[tex]F=\dfrac{\Delta p}{t}\\\\F=\dfrac{12.04}{0.012}\\\\=1003.33\ N[/tex]

Hence, this is the required solution.

Answer:

a. = 12.04 kg*m/s

b. = 1,003.3N

Explanation:

The answer above is correct.

A ball of mass m=10g, carrying a charge q =-20μe is suspended from a string of length L= 0.8m above a horizontal uniformly charged infinite plane sheet of charge density σ = 4μe/m^2. The ball is displaced from the vertical by an angle and allowed to swing from rest.

Required:
a. Obtain the equations of motion of the charged ball based on Newtonian laws of motion.
b. Assume the displaced angle θ is small and simplify the results obtained in part (a) to obtain the frequency of oscillations of the charged ball.

Answers

Answer:

a)       [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

b)     f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

Explanation:

a) To have the equations of motion, let's use Newton's second law.

Let's set a reference system where the x-axis is parallel to the path and the y-axis is in the direction of tension of the rope.

For this reference system the tension is in the direction of the y axis, we must decompose the weight and the electrical force.

Let's use trigonometry for the weight that is in the vertical direction down

             sin θ = Wₓ / W

             cos θ = W_y / w

             Wₓ = W sin θ

             W_y = W cos θ

we repeat for the electric force that is vertical upwards

              F_{ex} = F_e sin θ

              F_{ey} = F_e cos θ

the electric force is

               F_e = q E

where the field created by an infinite plate is

               E = [tex]\frac{ \sigma}{2 \epsilon_o}[/tex]  

let's write Newton's second law

Y  axis  

           T - W_y = 0

            T = W cos θ

X axis

            F_{ex} - Wₓ = m a                   (1)

           

we use that the acceleration is related to the position

            a = dv / dt

            v = dx / dt

where x is the displacement in the arc of the curve

substituting

            a = d² x /dt²

we substitute in 1

           q E sin θ - mg sin θ = m [tex]\frac{d^2 x}{dt^2}[/tex]

we have angular (tea) and linear (x) variables, if we remember that angles must be measured in radians

           θ = x / R

           x = R θ

we substitute

           sin θ (q E - mg) = m \frac{d^2 R \ theta}{dt^2}  

           

          [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] =  [tex]\frac{d^2 \theta}{d t^2}[/tex]

this is the equation of motion of the system

b) for small oscillations

         sin θ = θ

therefore the solution is simple harmonic

      θ = θ₀ cos (wt + Ф)

if derived twice, we substitute

- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o}  ) \frac{\theta}{R } θ₀ cos (wt + Ф) = -w² θ₀ cos (wt + Ф)

     

       w² =  [tex]\frac{g}{R}[/tex] - [tex]\frac{q}{m} \frac{ \sigma }{2 \epsilon_o} \frac{1}{R}[/tex]  

angular velocity is related to frequency

       w = 2π f

        f = 2π / w

        f = 2π/w

     

        f = 2π  [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]  

A ball is thrown straight up with an initial velocity of 128 ft/sec, so that its height (in feet) after t sec is given by s(t) = 128t-16t2. (a) What is the average velocity of the ball over the following time intervals? [3,4] ft/sec [3,3.5] ft/sec [3,3.1] ft/sec (b) What is the instantaneous velocity at time t = 3? ft/sec (c) What is the instantaneous velocity at time t = 6? ft/sec Is the ball rising or falling at this time? rising falling (d) When will the ball will hit the ground? t = sec

Answers

Answer:

one sec let me think

Explanation:

(a)The average velocity of the ball over the following time intervals will be  [3,4] ft/sec.

(b)The instantaneous velocity at time t = 3 will be32 ft/sec.

(c)The instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)The ball will hit the ground at 13.4 sec.

What is velocity?

The change of displacement with respect to time is defined as the velocity.  velocity is a vector quantity. it is a time-based component.

The given data in the question will be ,

u is the initial velocity by which ball thrown=128 ft/sec.

V₃ is the instantaneous velocity at time t=3 sec.

V₆ is the instantaneous velocity at time t=6 sec.

t is the time when ball hits the ground=?

(a) Given equation for the displacement

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

Time when velocity is zero will be

[tex]\rm{ t=\frac{128}{32}[/tex]

[tex]\rm{ t=4 sec[/tex]

If the velocity got in the equation is 128 and 32 ft /sec. it can be only when the average velocity is [3,4] ft/sec .

Hence the average velocity obtained from the problem will be  [3,4] ft/sec

(b)  

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=3 sec)

v(3) = 128-32×3

v(3) =32 m/sec.

Hence the instantaneous velocity at time t = 3 will be32 ft/sec.

(c)

s(t) =128t-16t²     (on differenting got the velocity )

v(t) = 128-32t

At time( t=6 sec)

v(6) = 128-32×6

v(6) = -64 m/sec.

Hence the instantaneous velocity at time t = 6 will be -64 ft/sec.

(d)

According to Newtons third equation of motion we got

v=u+gt

If the body returens from a certain height at max height its velocity must be zero; ( u=0)

[tex]\rm t=\frac{(v-u)}{g} \\\\\ \rm t=\frac{(128-0)}{9.81}\\\\\rm t=13.04 sec.[/tex]

Hence the ball will hit the ground at 13.4 sec.

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During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of 112 N/m. If the hose is stretched by 4.70 m and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length

Answers

Answer:

1237 J

Explanation:

The work done by the hose on the balloon is the work done by a spring which is

W = 1/2k(x₀² - x₁²) where k = spring constant = 112 N/m, x₀ = 4.70 m and x₁ = 0 m.

Substituting the values of the variables into the equation, we have

W = 1/2k(x₀² - x₁²)

W = 1/2 × 112 N/m((4.70 m)² - (0 m)²)

W = 56 N/m(22.09 m² - 0 m²)

W = 56 N/m(22.09 m²)

W = 1237.04 J

W ≅ 1237 J

Planets don't collide into
the sun because they

A. Are moving
B. Have too much mass
C. Have their own gravity
D. Are more attracted to each other

Answers

Every planet is at a different distance from the Sun and has a fixed orbit in which it revolves around the Sun. The Sun"s gravitational force holds the planets in this place and they do not collide with each other as their orbits are non-intersecting.
———
C. They have their own gravity

The local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.



Calculate the time required for bumper car A to travel the 10 m to collide with bumper car B.
Calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s.
Is the direction of motion for bumper car A following the collision with bumper car B to the right, to the left, or is bumper car A at rest?
Is the collision elastic? Justify your answer.

Answers

Answer:

a. 20 s

b. 0 m/s  

c. right

d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right

Explanation:

im not sure though

By applying conservation of linear momentum, the answers are:

1. Time = 2 s

2. 3 m/s

3. same direction

4. Inelastic collision

COLLISION

There are for types of collision. They are;

Elastic CollisionPerfectly elastic collisionInelastic collisionPerfectly Inelastic collision

Given that a local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.

1. The time required for bumper car A to travel the 10 m to collide with bumper car B can be calculated by using first equation of linear motion.

V = U + at

Where

V = 4 m/s

U = 2 m/s

a = 1 m/[tex]s^{2}[/tex]

Substitute all the parameters into the formula

4 = 2 + t

t = 4 - 2

t = 2s

2. To calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s, we will apply conservation of linear momentum

[tex]m_{1}u_{1}[/tex] = [tex]m_{1}v_{1}[/tex] + [tex]m_{2}v_{2}[/tex]

300 x 4 = 300V + 200 x 3

1200 = 300V + 300

300V = 1200 - 300

300V = 900

V = 900/300

V = 3 m/s

3. Since the final velocity of car A is positive, the direction of motion for bumper car A follows the collision with bumper car B to the right.

4. Since the both move at the same velocity, the collision inelastic.

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Explain which energies work together to bake the bread
and cook the eggs.

Answers

Answer:

Thermal Energy

Explanation:

The energies that work together to bake the bread and cook the eggs is the thermal energy. Thermal energy is basically heat energy, which makes the food warmer.

Thermal energy and chemical energy work together to bake the bread and cook the eggs.

What is thermal energy?

The energy present in a system that determines its temperature is referred to as thermal energy. Thermal energy flows as heat. Thermodynamics is a whole field of physics that studies how heat is transmitted across various systems and how work is done in the process.

What is chemical energy?

Chemical energy is described as: the power that is kept in chemical compound bonds (molecules and atoms). It is released during the chemical process, which is referred to as an exothermic reaction, which mostly generates heat as a byproduct.

When the bread is baked or the eggs are cooked, thermal energy is provided from outside and  chemical energy inside the molecules of bread  or eggs come to play. So, these two energies work together to bake the bread and cook the eggs.

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PLEASE HELP ME WITH THIS ONE QUESTION

Which pendulum would have the highest frequency?

1) 1 kg mass and 0.5-meter length

2) 0.5 kg mass and 0.75-meter length

3) 0.5 kg mass and 1-meter length

4) 1 kg mass and 0.75-meter length

Answers

Answer:

B is the Answer

Follow me please

Mark brainliest

Answer:

The answer is A

Explanation:

This is because f = 1/T,

or frequency equals 1 over period, T

Since the highest T would be the answer with the greatest length and mass does not matter the answer would be c IF they were asking for the highest period.

F is found by 1 over T so the greatest F would be the smallest number, Hence answer A

Please Mark me brainiest

weight is measured in units called

Answers

Pounds

If you are talking about the unit of measurement for weight is that of force it would be Newtons.

The weight is measured in a unit called Newton. The newton is a standard unit of weight measurements.

The International System of Units (SI) uses Newtons (N) as the unit of weight measurement. Weight is frequently expressed in pounds (lb) or ounces (oz) in some nations that employ the Imperial system. However, Newton is the accepted weight measurement unit in scientific and international contexts.

The International System of Units (SI) uses Newton as the primary unit of weight measurement. Sir Isaac Newton, a great physicist who significantly influenced our understanding of classical mechanics, is honored by having his name attached to it. Newton is the force needed to accelerate a mass of one kilogram by one meter per second squared in the SI system.

Hence, the weight is measured in a unit called Newton. The newton is a standard unit of weight measurements.

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Which statement best describes this situation

Answers

Answer:

what situation?

Explanation:

You are driving your car on a very cold late Fall day. You clear a turn and see a couple of pedestrians standing at the cross walk. They are eager to cross the road and to get into the warmth of their apartment as soon as possible. You have two options: continue driving your car as you were without lowering your speed and drive right by the pedestrians OR slow down, stop right at the crosswalk, and yield to the pedestrians. Although by Virginia law the choice is clear, what about Physics laws? Which scenario (passing by or slowing down and stopping at the crosswalk to yield) will minimize the time the pedestrians are out in the cold freezing before they can cross the road?
Make the following assumptions in your argument. Before you noticed the pedestrians, you are moving with a constant velocity v=22 miles/hour. The distance at which you noticed the pedestrians is D=23 meters. Write down a symbolic expression for the amount of time, tpass , the pedestrians will have to wait till they cross the road if you simply drive by without slowing down or speeding up.
Write down a symbolic expression for the amount of time, tstop, the pedestrians will have to wait till they cross the road if you slow down, come to a complete stop at the crosswalk and yield to the pedestrians.

Answers

Answer:

t_pass = 2.34 m

t_stop = 4.68 s

Thus, for the car passing at constant speed the pedestrian will have to wait less.

Explanation:

If the car is moving with constant speed, then the time taken by it will be given as:

[tex]t_{pass} = \frac{D}{v}[/tex]

where,

t_pass = time taken = ?

D = Distance covered = 23 m

v = constant speed = (22 mi/h)(1609.34 m/1 mi)(1 h/3600 s) = 9.84 m/s

Therefore,

[tex]t_{pass} = \frac{23\ m}{9.84\ m/s} \\[/tex]

t_pass = 2.34 m

Now, for the time to stop the car, we will use third equation of motion to get the acceleration first:

[tex]2as = v_{f}^{2} - v_{i}^2\\a = \frac{v_{f}^{2} - v_{i}^2}{2D}\\\\a = \frac{(0\ m/s)^{2}-(9.84\ m/s)^2}{(2)(23\ m)}\\\\a = -2.1\ m/s^2[/tex]

Now, for the passing time we use first equation of motion:

[tex]v_{f} = v_{i} + at_{stop}\\t_{stop} = \frac{v_{f}-v_{i}}{a}\\\\t_{stop} = \frac{0\ m/s - 9.84\ m/s}{-2.1\ m/s^2}[/tex]

t_stop = 4.68 s

Constant velocity is the velocity which covers the same distance for each interval of the time.

The time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

What is constant velocity?

Constant velocity is the velocity which covers the same distance for each interval of the time.

It can be given as,

[tex]v=\dfrac{x}{t}[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.

Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.

Thus amount of time, [tex]t_{pass}[/tex] is,

[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]

According to the third equation of the motion acceleration can be given as,

[tex]v^2-u^2=2ax\\a=\dfrac{v^2-u^2}{2x}\\a=\dfrac{0^2-9.84^2}{2\times 23}\\a=-2.1 \rm \; m/s^2[/tex]

Now, use the first equation of motion, to get the required time,

[tex]v=u+at\\0=9.84+(-2.1)t\\t=4.68\rm \; s[/tex]

Therefore, the time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.

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Please answer :>
40 POINTS

Answers

Answer:

rotates

Explanation:

I'm so bored

yrfgggghhgghhyuj

Rotates because it’s on a certain degree

3. If a spring extends by 3 cm when a 4 N weight is suspended from it, find the extension
when the weight is changed to
(a) 8 N
(b) 10 N
(c) 14 N

Answers

8N - 6cm
10N - 7.5cm
14N - 10.5cm

Waves are ?
that can travel through matter.

Answers

Answer:

A wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.

What is the gravitational potential energy of an object that has a mass of 8 kg and is 11.2 meters above Earth? Round your answer the nearest whole number.
A. 878 J
B. 30 J
C. 680 J

Answers

Answer:
A.
Explanation:
GPE= ham
GPE= (8)(11.2)(9.8)
GPE= 878.08
round it up
878 J

PLEASE HELP!!!



What are some compound machines that are used today?

Answers

Answer:

Examples of compound machines include bicycles, cars, scissors, and fishing rods with reels. Compound machines generally have lower efficiency but greater mechanical advantage than simple machines

Brilianst

Answer:

screws, inclined planes , etc

Explanation:

no explanation needed

A screw-jack used to lift a bus is a
A) first order lever
B) second order lever
C) pulley
D) screw

Answers

Answer:

c

Explanation:

One of the smallest planes ever flown was the Bumble Bee II, which had a mass of 180 kg. If the pilot’s mass was 70 kg, what was the velocity of both plane and pilot if their momentum was 20,800 kg∙m/s to the west?

Answers

Answer:

83.2 m/s to the West

Explanation:

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

Mass of plane = 180 Kg

Mass of pilot = 70 Kg

Momentum = 20800 Kg∙m/s West

Velocity =?

Next, we shall determine the total mass. This can be obtained as follow:

Mass of plane = 180 Kg

Mass of pilot = 70 Kg

Total mass =?

Total mass = Mass of plane + Mass of pilot

Total mass = 180 + 70

Total mass = 250 Kg

Finally, we shall determine the velocity. This can be obtained as follow:

Total mass = 250 Kg

Momentum = 20800 Kg∙m/s West

Velocity =?

Momentum = mass × Velocity

20800 = 250 × Velocity

Divide both side by 250

Velocity = 20800 / 250

Velocity = 83.2 m/s West

Thus, the velocity of both plane and pilot is 83.2 m/s to the West

A dump truck, whose bed is made of steel, holds an old steel watering trough. The bed of the truck is slowly raised until the trough begins to slide. For dry steel to steel μs= 0.80, μk= 0.60. What is the acceleration of the trough as it slides down the truck bed? Express your answer with the appropriate units.

Answers

Answer:

a = 1,538 m / s²

Explanation:

Let's use Newton's second law, let's set a reference system where the x-axis is parallel to the sloping floor of the truck and the positive direction is in the direction of movement of the trough, for this case the weight is the only force to decompose

          sin θ = Wₓx / W

          cos θ = W_y / W

          Wₓ = W sin θ

          W_y = W cos θ

Y axis

          N -W_y = 0

          N = mg cos θ

X axis

          Wₓ - fr = m a

the friction force has the expression

          fr = μ N

There are values ​​of the friction coefficient (μ_s) one for when the movement has not started and it takes a smaller value for when the bodies are moving.

In this case we first find the angle for which the movement begins, in this part we use the static coefficient and the acceleration is zero

             Wₓ - μ_s N = 0

             m g sin θ = μ_s mg cos θ

             tan θ = μ_s

             θ = tan⁻¹ μ_s

we calculate

              θ = tan⁻¹ 0.8

              θ = 38.7º

For this angle, how the trough begins to move, the coefficient is reduced to the dynamics coefficient (μ_k)  and the acceleration is different from zero.

     

we substitute

          mg sin θ - μ_k mg cos θ = m a

          a = g (sin θ - μ_k cos θ)

let's calculate

          a = 9.8 (sin 38.7 - 0.6 cos 38.7)

          a = 1,538 m / s²


The main difference between a chest and a bounce pass is what?

Answers

Answer: The main difference between the three is the mode of transmission. The chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate. Finally, the overhead pass is projected high in the air to avoid defenders.

Explanation:

Answer:

The main difference is: the chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate.

The secondary main difference is the amount of power from the ball recived from the person reciving

What is the wavelength of this wave

Answers

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the Concept of the Waves.

Wavelength is the distance between two consecutive crest or trough.

hence, here the distance is 10cm

So the wavelength is 10cm

===> 10 cm

Why is fluorine more chemically reactive than nitrogen

Answers

Answer:

Me encanta el pollo, de todos modos, un día que es hoy, por cierto, fui a la tierra de los pollos y caminé en México, ¿sabías que es mi sueño? ¿Olvidé que mi nombre es jemma y me gusta caminar? párrafo así que nos vemos

Explanation:

Answer:

Fluorine needs to gain only 1 electron to have a full octet, while nitrogen needs to gain 3.

What is the weight of a 25 kg object on Earth with an acceleration due to gravity of 9.8m/s/s?

2.45 n
24.5 n
245 n
2450 n​

Answers

245 n hope this helps

Which word best completes the sentence?

Select the word from the drop-down menu

He is quite
Choose...
despite never having left his smalL TOWEN

Answers

Answer:

it’s cosmopolitan

Explanation:

k12

Suppose that a uniform rope of length L resting on a frictionless horizontal surface, is accelerated along the direction of its length by means of a force F, pulling it at one end. A mass M is accelerated by the rope. Assuming the mass of the rope to be m and the acceleration is a. Stated in terms of the product ma, what is the tension in the rope at the position 0.3 L from the end where the force F is applied if the mass M is 1.5 times the mass of the rope m?

Answers

Answer:

2.2 ma

Explanation:

Given :

Length of the rope = L

Mass of the rope = m

Mass of the object pulled by the rope = M

M = 1.5 m

So, L [tex]$\rightarrow$[/tex] m

For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]

∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.3 m

And for remaining 0.7 L =  [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]

            = 0.7 m

By Newtons law of motion,

F - T = ( 0.3 m) a .........(1)

T = ( M + 0.7 m) a

T = ( 1.5 m + 0.7 m) a

T = ( 2.2 m ) a  ..............(2)

So from equation (1) and (2), we have

Tension on the rope

T = 2.2 ma

Please help me with this 29 points

Answers

Answer:

)Give the definition of poverty line as defined by the World Bank.

6xy from -12xy
please give me a answer this question​

Answers

6 floors down from 12 floors underground = 18 floors underground.

6 degrees colder than 12 degrees below zero  =  18 degrees below zero

6 brown cows taken away from -12 brown cows = -18 brown cows

6 cars sold from a dealer that 12 cars were stolen from = 18 cars gone

6xy taken away from -12xy  =  -18xy

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