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BARSIC [14]
3 years ago
13

6. A 4 kg object hangs below a 6 kg object by a string of negligible mass. If the 6 kg object is pulled upward by a force of 440

N, what is the tension in the string?

Physics
1 answer:
MrRissso [65]3 years ago
7 0

Answer:

T =176 N

Explanation:

from diagram

F -(m_1+m_2_g) = (m_1+m_2_g)a

440 - (6+4)g = (6+4)a

a =\frac{440-10*9.8}{10}

a =34.2 m/s^2

frrom free body diagram of mass m2 = 4kg

T -m_2g =m_2a

T = m_2(g +a)

T = 4(9.81+34.2)

T =176 N

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A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un
Kamila [148]

Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c)  </em>\frac{4492.18}{v_{car} ^{2} }

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

I = \frac{1}{2}mr^{2}

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

I =  \frac{1}{2}*11*1.1^{2} = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = Iw^{2} = 6.655 x 31.82^{2} = <em>6738.27 J</em>

<em></em>

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

I = \frac{1}{2}mr^{2} =  \frac{1}{2}*16*2.8^{2} = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = Iw = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

(I_{1} +I_{2} )w

where the subscripts 1 and 2 indicates the values first and second  flywheels

(I_{1} +I_{2} )w = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = Iw^{2} = 6.655 x 3.05^{2} = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = \frac{1}{2}mv_{car} ^{2}

where m is the mass of the car

v_{car} is the velocity of the car

Equating the energy

2246.09 =  \frac{1}{2}mv_{car} ^{2}

making m the subject of the formula

mass of the car m = \frac{4492.18}{v_{car} ^{2} }

3 0
3 years ago
A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra
givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a

divide both sides by 40

a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2

5 0
3 years ago
A guitar vibrates in frequency with a tuning fork when the fork is held against its body. This is a case of
Butoxors [25]
Energy transfer the energy from the tuning fork is being transferred to the guitar<span />
4 0
3 years ago
You want to make a ride so you do not want to exceed 1.1g’s, if the radius of the turns are 10m, then what is the maximum speed
Citrus2011 [14]

The maximum speed is 10.4 m/s

Explanation:

For a body in uniform circular motion, the centripetal acceleration is given by:

a=\frac{v^2}{r}

where

v is the linear speed

r is the radius of the circular path

In this problem, we have the following data:

- The maximum centripetal acceleration must be

a=1.1 g

where g=9.8 m/s^2 is the acceleration of gravity. Substituting,

a=(1.1)(9.8)=10.8 m/s^2

- The radius of the turn is

r = 10 m

Therefore, we can re-arrange the equation to solve for v, to find the maximum speed the ride can go at:

v=\sqrt{ar}=\sqrt{(10.8)(10)}=10.4 m/s

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

8 0
3 years ago
Calculate the density of a material that has a bat a mass of 52.457 g any volume of 13.5 cm³
3241004551 [841]
Answer:
d = 3.8857 g/cm^3

Explain:
Formula/: d = m/V

Hope this helps!
(a brainliest would be appreciated)
4 0
3 years ago
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