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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 2.3 kg block of copper is heated at atmospheric pressure such that its temperature increases from 6 oC to 90 oC. How much heat
Shtirlitz [24]

Answer:

the  heat absorbed by the block of copper is 74368.476J

Explanation:

Hello!

To solve this problem use the first law of thermodynamics that states that the heat applied to a system is the difference between the initial and final energy considering that the mass and the specific heat do not change so we can infer the following equation

Q=mCp(T2-T1)

Where

Q=heat

m=mass=2.3kg

Cp=0.092 kcal/(kg C)=384.93J/kgK

T2=Final temperatura= 90C

T1= initial temperature=6 C

solving

Q=(2.3kg)(384.93\frac{J}{kgC} )(90C-6C)=74368.476J

the  heat absorbed by the block of copper is 74368.476J

7 0
3 years ago
A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth =  5.98 × 10²⁴ kg

Radius of Earth = 6.38\times 10^6\ m

The gravitational force will balance the centripetal force

\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}

T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}

From Kepler's law we have relation

T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m

Distance from the center of the Earth would be

42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}

8 0
3 years ago
a ball at rest starts rolling down a hill with a constant acceleration of 3.2 meters/second2. what is the final velocity of the
Y_Kistochka [10]

Answer:

3.2(6.0) = 19.2 m/s

Explanation:

6 0
3 years ago
The atmosphere is held together by
Vera_Pavlovna [14]

Answer:

D. gravity

Explanation:

Gravity keeps the atmosphere from escaping into space.

3 0
3 years ago
suppose you have a 69.0-kg wooden crate resting on a wood floor. what maximum force can you exert horizontally on the crate with
marusya05 [52]

You've got a 69.0-kg wooden crate on a wooden floor. The box can withstand a force of up to 338N in a horizontal direction without being moved. Following this, the wooden creates moving stats.

In order to calculate the friction coefficient, divide the force pushing two objects together by the force acting between them. friction coefficient might be 0 or one. They can be split into two categories: friction coefficient that is static. Kinetic friction coefficient (also known as sliding coefficient of friction).

the acceleration brought on by the gravitational pull of large masses generally, gravitational , often known as the acceleration brought on by the Earth's gravitational pull and centrifugal force,

F= friction coefficient *M*g

F= 0.5*69*9.8

F=338N

Learn more about gravitational here

brainly.com/question/3009841

#SPJ4

7 0
1 year ago
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