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kondor19780726 [428]
2 years ago
12

A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the fi

rst complete revolution, how long will it take to make the next complete revolution?
Physics
1 answer:
miv72 [106K]2 years ago
7 0

Answer:

The time taken is  \Delta t  = 1.5 \ s

Explanation:

From the question we are told that

   The mass of the ball is  m =  1.25 \ kg

    The time taken to make the first complete revolution is  t= 3.60 s

    The displacement of the first complete revolution is  \theta  =  1 rev  =  2 \pi \  radian

Generally the displacement for one  complete revolution is mathematically represented as

       \theta =  w_i t  +  \frac{1}{2} *  \alpha  * t^2

Now given that the stone started from rest w_i  = 0 \ rad / s

     2 \pi =0   +  0.5*  \alpha  *(3.60)^2

     \alpha   =  0.9698 \  s

Now the displacement for two  complete revolution is

         \theta_2  =  2 *  2\pi

         \theta_2  = 4\pi

Generally the displacement for two complete revolution is mathematically represented as  

     4 \pi =   0  +  0.5 * 0.9698 * t^2

=>   t^2  =  25.9187

=>   t=  5.1 \ s

So

 The  time taken to complete the next oscillation is mathematically evaluated as

     \Delta t  =  t_2  - t

substituting values

      \Delta t  = 5.1 -  3.60

     \Delta t  = 1.5 \ s

           

 

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Answer:

The minimum speed = \sqrt{\frac{2GM}{R} }

Explanation:

The minimum speed that the rocket must have for it to escape into space is called its escape velocity. If the speed is not attained, the gravitational pull of the planet would pull down the rocket back to its surface. Thus the launch would not be successful.

The minimum speed can be determined by;

                      Escape velocity = \sqrt{\frac{2GM}{R} }

where: G is the universal gravitational constant, M is the mass of the planet X, and R is its radius.

If the appropriate values of the variables are substituted into the expression, the value of the minimum speed required can be determined.

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3 years ago
What is the pendulum length whose period is 2.0s ?
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 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
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Which waves can travel through space?
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Explanation:

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You are rushing to the train station to catch your morning commute. The train leaves the train station from rest with an acceler
Vanyuwa [196]

Answer:

a) t= 4.81 s,and t= 23.51.

b)d=6.88 m

c)v=4.8 m/s

Explanation:

Acceleration of train ,a= 0.6 m/s²

u = 0 m/s

Your speed ,V= 8.5 m/s

Lets take after t time he you catch the train

Distance travel by train in t time

d=ut+\dfrac{1}{2}at^2

d=\dfrac{1}{2}\times 0.6\times t^2  ----------1

d= V ( t- 4)

d= 8.5 ( t- 4)  --------2

By equating equation 1 and 2

0.3 t² =  8.5 ( t- 4)

0.3 t² -8.5 t + 34 = 0

t= 4.81 s and t= 23.51

It means that you will catch train after t= 4.81 s,and t= 23.51.

t=23.51 sec means that you will catch the train after 23.51 sec also because acceleration of train is low.

Distance travel

d= 8.5 ( t- 4)

t= 4.81 s

d= 8.5 ( 4.81- 4)  m

d=6.88 m

Lets speed = v

0.3 t² =  v ( t- 4)

0.3 t² - v t + 4 v = 0

To have one solution

D=\sqrt{b^2-4ac}

D= 0 Should be zero.

v²- 4 x 0.3 x 4 v

v = 4 x 0.3 x 4

v=4.8 m/s

6 0
3 years ago
How much heat is released upon converting one mole of steam (18.0
Anika [276]

4.65 × 10⁴ Joules of heat is released upon converting one mole of steam to water.

\texttt{ }

<h3>Further explanation</h3>

Specific Heat Capacity is the amount of energy needed to raise temperature of 1 kg body for 1°C.

\large {\boxed{Q = m \times c \times \Delta t} }

<em>Q = Energy ( Joule )</em>

<em>m = Mass ( kg ) </em>

<em>c = Specific Heat Capacity ( J / kg°C ) </em>

<em>Δt = Change In Temperature ( °C )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

initial temperature of steam = t = 100.0°C

specific heat capacity of water = c = 4.186 J/gK

mass of steam = m = 18.0 gram

final temperature of water = t' = 25.0°C

specific latent heat of vaporization of water = Lv = 2268 J/g

<u>Asked:</u>

heat released = Q = ?

<u>Solution:</u>

\boxed {\large {\texttt{steam 100}^oC \overset{[Q_1]}{\rightarrow} \texttt{water 100}^oC \overset{[Q_2]}{\rightarrow} \texttt{water 25}^oC}}

Q = Q_1 + Q_2

Q = mL_v + mc\Delta t

Q = 18.0(2268) + 18.0(4.186)(100-25)

Q = 40824 + 5651.1

Q = 46475.1 \texttt{ Joules}

Q \approx 4.65 \times 10^4 \texttt{ Joules}

\texttt{ }

<h3>Learn more</h3>
  • Efficiency of Engine : brainly.com/question/5597682
  • Flow of Heat : brainly.com/question/3010079
  • Difference Between Temperature and Heat : brainly.com/question/3821712

\texttt{ }

<h3>Answer details </h3>

Grade: College

Subject: Physics

Chapter: Thermal Physics

\texttt{ }

Keywords: Heat , Temperature , Block , Aluminium , Ice , Cold , Water

4 0
3 years ago
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