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3 years ago

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A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th

en what is the speed of ball

1 answer:

kondaur [170]3 years ago

5 0

The speed of the ball is 1.57 m/s

Explanation:

We start by calculating the angular speed of the ball, which is given by

$\omega = \frac{2\pi}{T}$

where

T is the period of revolution

For the ball in this problem,

T = 2 s

(since the ball makes 1 revolution every 2 seconds)

So, the angular speed is

$\omega = \frac{2\pi}{T}=\frac{2\pi}{2}=\pi rad/s$

Now we can calculate the speed of the ball by using

$v=\omega r$

where

r = 0.5 m is the radius of the circular path

Substituting,

$v=(\pi)(0.5) = 1.57 m/s$

Learn more about angular motion:

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2 years ago

If the water in the bathtub has a higher temperature than the air in the bathroom, which will occur?

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Heat will flow from the water into the air by convection .

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3 years ago

As the collision frequency of gas particles increases, the mean free path of the gas particles ____

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As the collision frequency of gas particles increases, the mean free path of the gas particles decreases.

<h3>Frequency </h3>

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2 years ago

Question 2 (ID=81813)

zaharov [31]

Answer:

The answer is B

Explanation:

5/2=2.5

2.5x2=5

Hope this helps ik its kinda confusing lol

7 0

3 years ago

Read 2 more answers

An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,

Sedbober [7]

Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

a. Internal energy and the relative specific volume at $s_1$ are determined from A-17: $u_1=214.07kJ/kg, \ \alpha_r_1=621.2$ .

The relative specific volume at $s_2$ is calculated from the compression ratio:

$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$

Relative SV at $s_4$ is

$a_r_4=\frac{r}{r_c}\alpha _r_3$

= $=\frac{16}{2}\times4.533\\=36.424$

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

3 0

3 years ago