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Norma-Jean [14]
3 years ago
12

A ball of mass 1.6 kg is attached to the end of a massless string. A circus clown twirls the

Physics
1 answer:
erik [133]3 years ago
7 0

Compute the ball's angular speed <em>v</em> :

<em>v</em> = (1 rev) / (2.3 s) • (2<em>π</em> • 180 cm/rev) • (1/100 m/cm) ≈ 4.917 m/s

Use this to find the magnitude of the radial acceleration <em>a</em> :

<em>a</em> = <em>v </em>²/<em>R</em>

where <em>R</em> is the radius of the circular path. We get

<em>a</em> = <em>v</em> ² / (180 cm) = <em>v</em> ² / (1.8 m) ≈ 13.43 m/s²

The only force acting on the ball in the plane parallel to the circular path is the tension force. By Newton's second law, the net force acting on the ball has magnitude

∑ <em>F</em> = <em>m</em> <em>a</em>

where <em>m</em> is the mass of the ball. So, if <em>t</em> denotes the magnitude of the tension force, then

<em>t</em> = (1.6 kg) (13.43 m/s²) ≈ 21 N

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510 g squirrel with a surface area of 935 cm2 falls from a 4.8-m tree to the ground. Estimate its terminal velocity. (Use the dr
Yuki888 [10]

Answer:

The terminal velocity is v_t  =17.5 \ m/s

Explanation:

From the question we are told that

       The mass of the squirrel is  m_s  =  50\ g  =  \frac{50}{1000} =  0.05 \  kg

      The surface area is   A_s =  935 cm^2  =  \frac{935}{10000} = 0.0935 \ m^2

       The height of fall is  h =4.8 m

        The length of the prism is l =  23.2 = 0.232 \ m

          The width of the prism is w =  11.6 =  0.116 \ m

 

The terminal velocity is mathematically represented as

       v_t  =  \sqrt{\frac{2 * m_s *  g }{\dho_s * C  * A } }

Where \rho  is the density of a rectangular prism with a constant values of \rho  =  1.21 \ kg/m^3

            C is the drag coefficient for a horizontal skydiver with a value = 1

            A  is the area of the prism the squirrel is assumed to be which is mathematically represented as

      A =  0.116 * 0.232

       A =  0.026912 \ m^2

 substituting values

      v_t  =  \sqrt{\frac{2 * 0.510 *  9.8 }{1.21 * 1  * 0.026912 } }

     v_t  =17.5 \ m/s

       

7 0
4 years ago
Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the
zlopas [31]

Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

\rho = Density of medium

A = Area

\nu = Frequency

\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

8 0
3 years ago
How does the speed of the different molecules that make up the air depend on their masses?
koban [17]
Larger molecules will move slower and smaller molecules will move faster. Did this answer your question?

8 0
3 years ago
"It is not correct to say that a body contains a certain amount of heat, yet a body can transfer heat to another body. How can a
saveliy_v [14]

Answer:

Because heat is a path function or the energy in transit.

Explanation:

  • It is not correct to say that a body contains a certain amount of heat because the heat is a path function and not a property of the system. It is the energy in transit which can be encountered only when it crosses the system boundary.
  • Heat is the energy in transit of a matter which flows by the virtue of temperature difference. The heat energy in a body is stored in the form of kinetic energy of the molecules which gets converted into heat that we know as the responsible factor for the rise in temperature usually.
5 0
4 years ago
011 10.0 points
Sedbober [7]
<h2>The temperature of the air is 66.8° C</h2>

Explanation:

From the Newton's  velocity of sound relationship , the velocity of sound is directly proportional to the square root of temperature .

In this case The velocity of sound = frequency x wavelength

= 798 x 0.48 = 383 m/sec

Suppose the temperature at this time = T K

Thus 383 ∝ \sqrt{T}                       I

The velocity of sound is 329 m/s at 273 K ( given )

Thus 329 ∝ \sqrt{273}                   II

Dividing I by II , we have

\frac{383}{329} = \sqrt{\frac{T}{273} }

or \frac{T}{273} = 1.25

and T = 339.8 K  = 66.8° C

4 0
4 years ago
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