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Alik [6]
3 years ago
5

A dad pushes tangentially on a small hand-driven merry-go-round and is able to accelerate it from rest to a frequency of 18 rpm,

in 13.0 s. Assume the merry-go-round is a uniform disk of radius 2.7 m; and has a mass of 800 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. Calculate the torque required to produce the acceleration, neglecting frictional torque. What force is required at the edge?
Physics
1 answer:
puteri [66]3 years ago
4 0

Answer:

F = 176.175 N

Explanation:

given,  

radius of merry - go - round = 2.7 m

mass of the disk = 800 kg  

speed of the merry- go-round = 18 rpm  

                                                 = 18 \dfrac{2\pi }{60}

                                                 = 1.88 rad/s

time = 13 s

mass of two children = 25 kg  

ω = ω₀ + α t

1.88 = 0 + α(13)

α = 0.145 rad/s²

we know

τ = I α

so,

I = \dfrac{1}{2}MR^2+ 2 MR^2  

I = \dfrac{1}{2}\times 800 \times 2.7^2+ 2\times 25\times 2.7^2

I = 3280.5 kg.m^2  

τ = I α

τ = 3280.5 x 0.145

τ = 475.67 N m

τ = F x r

475.67 = F x 2.7

F = 176.175 N

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Train cars are coupled together by being bumped into one another. Suppose two loaded cars are moving toward one another, the fir
tia_tia [17]

Answer:

7560 Joules

Explanation:

m_1 = Mass of first car = 1.5\times 10^5\ kg

m_2 = Mass of second car = 2\times 10^5\ kg

u_1 = Initial Velocity of first car = 0.3 m/s

u_2 = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

As linear momentum of the system is conserved

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{1.5\times 10^5\times 0.3 + 2\times 10^5\times -0.12}{1.5\times 10^5 + 2\times 10^5}\\\Rightarrow v=0.06\ m/s

Energy lost is

\Delta E=\Delta E_i-\Delta E_f\\\Rightarrow \Delta=\frac{1}{2}(m_1u_1^2 + m_2u_2^2-(m_1+m_2)v^2)\\\Rightarrow \Delta=\frac{1}{2}(1.5\times 10^5\times 0.3^2 + 2\times 10^5\times (-0.12)^2-(1.5\times 10^5 + 2\times 10^5)\times 0.06^2)\\\Rightarrow \Delta=7560\ J

The Energy lost in the collision is 7560 Joules

7 0
3 years ago
if an object is moving at a constant speed in a changing direction, What is the acceleration of the object (zero or not zero)​
Andreas93 [3]
  • We know, acceleration is the change of velocity by time.
  • Velocity is the speed of an object which also indicates the direction.
  • Hence, acceleration is both dependant upon the speed as well as the direction.
  • So, if an object is moving at a constant speed in a changing direction, the acceleration will also change. It will not be zero.
  • An example is that of uniform circular motion.

Answer:

if an object is moving at a constant speed in a changing direction, the acceleration of the object will not be zero.

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Hi I need help with some questions I have on a quiz / test
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3 years ago
What is the acceleration of a dog that runs from 3 m/s to 6 m/s over a distance of 90m?
KonstantinChe [14]

Answer:

solution given:

acceleration (a)=?

initial velocity (u)=3m/s

final velocity (v)=6m/s

distance (s)=90m

we have

v²=u²+2as

substituting value

6²=3²+2*a*90

36=9+180a

36-9=180a

a=25/180

<u>a=0.1388m/s²</u>

6 0
3 years ago
table 4.interaction between two permanent bar magnets what i did to the pair of magnets to cause interaction and observed the ef
Leni [432]

A magnet is a substance which attracts or repels another substance. In a magnet, the atoms are aligned in a particular direction in domains. A magnet has two poles: North pole and South pole. The domains are oppositely aligned in unlike poles. Like poles repel each other where as unlike poles attract each other. Hence, when we bring like poles closer, repulsion would be experienced. In case of unlike poles, they would stick together.

6 0
3 years ago
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