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Lelechka [254]
3 years ago
14

A 110-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

bout the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s
Physics
1 answer:
sweet [91]3 years ago
8 0

Answer:

155 N

Explanation:

We are given following data for a uniform, solid, horizontal disk:  

r = 1.5 m

m= 110 kg

t=2 s

w = 0.6  

Torque is given by:  

т = F*r

  =I*α

Solving it for the force exerted on the rope:  

F = I*α/r

  = (1/2*m*r^2)*(2π*w/t )/r

  = (1/2*110*1.5^2)*(2π*0.6/2 )/1.5

  = 155 N

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A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with
Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = \dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2

              = \dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2

              = 31 J

Final KE = \dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0
3 years ago
A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th
ale4655 [162]

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that  opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

     -Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

 a is the acceleration of the car

    a = \frac{frictional force}{mass} = \frac{10500}{1500} = -7m/s²

Now using;

   V² = U² + 2as

   V is the final velocity

   U is the initial velocity

   a is the acceleration

   s is the distance moved

  0² = 25² + 2 x 7 x s

  0 = 625 - 14s

  -625 = -14s

      s = 44.64m

   

learn more:

Velocity problems brainly.com/question/10932946

#learnwithBrainly

7 0
3 years ago
Adita lifts a book from the floor, carries it across the room, and places it on a high shelf. When is Adita doing work on the bo
noname [10]

She does work from the moment she touches the book until she lets it go. Work is anything that requires energy. Therefore, she is working as she picks up the book, carries it, and when she is lifting it onto the shelf.

5 0
3 years ago
Read 2 more answers
Suppose a car approaches a hill and has an initial speed of 108 km/h at the bottom of the hill. The driver takes her foot off of
Aleks04 [339]

Answer:

a) The car will reach a height of 45.9 m.

b) The amount of thermal energy generated is 173382 J.

c) The magnitude of the force of friction is 417.8 N.  

Explanation:

Hi there!

a) In this problem, we have to use the conservation of energy. The energy conservation theorem states that the energy of a system remains constant. Energy can´t be created nor destroyed, only transformed. In the case of the car, the initial kinetic energy is transformed into potential energy as the car´s height increases while coasting up the hill.

Then, all the initial kinetic energy (KE) will be transformed into potential energy (PE) (only if there is no friction).

The equation of KE is the following:

KE = 1/2 · m · v²

Where:

m = mass of the car.

v = speed of the car.

The equation of PE is the following:

PE = m · g · h

Where:

m = mass of the car.

g = acceleration due to gravity.

h = height at which the car is located.

Since work done by friction is negligible, we can assume that all the initial kinetic energy will be transformed into potential energy. Then:

KE at the bottom of the hill = PE at the top of the hill

1/2 · m · v² = m · g · h

Solving for h:

1/2 · v² / g = h

Let´s convert the speed unit into m/s:

108 km/h · 1000 m/ 1 km · 1 h / 3600 s = 30 m/s

Now, let´s calculate h:

h = 1/2 · (30 m/s)² / 9.8 m/s²

h = 45.9 m

The car will reach a height of 45.9 m.

b) In this case, all the kinetic energy is not transformed into potential energy because some energy is transformed into thermal energy due to friction. The thermal energy generated is equal to the work done by friction. Then:

KE at the bottom of the hill = PE + work done by friction

KE = PE + Wfr  (where Wfr is the work done by friction).

1/2 · m · v² = m · g · h + Wfr

1/2 · m · v² - m · g · h = Wfr

1/2 · 710 kg · (30 m/s)² - 710 kg · 9.8 m/s² · 21 m = Wfr

Wfr = 173382 J

The amount of thermal energy generated is 173382 J.

c) The work done by friction is calculated as follows:

Wfr = Ffr · Δx

Where:

Ffr = friction force.

Δx = traveled distance

Please, see the attached figure to notice that the traveled distance can be calculated by trigonometry using this trigonometric rule of right triangles:

sin angle = opposite side / hypotenuse

In our case:

sin 2.9° = h / Δx

Δx = h / sin 2.9°

Δx = 21 m / sin 2.9° = 415 m

Then, solving for the friction force using the equation of the work done by friction:

Wfr = Ffr · Δx

Wfr / Δx = Ffr

173382 J / 415 m = Ffr

Ffr = 417.8 N

The magnitude of the force of friction is 417.8 N

6 0
3 years ago
A 150 kg object and a 450 kg object are separated by 0.430 m. (a) find the net gravitational force exerted by these objects on a
Dvinal [7]
Forces are vectors. They cancel each other when in opposite directions. Use the equation F=GMm/r^2 and find the difference between the two forces since they are directly opposing each other.
4 0
3 years ago
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