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3 years ago

Picture a ball traveling at a constant speed around the inside of a circular Structure. is the ball acceleration? Explain why?

2 answers:

8 0

Yes. On a circular path, the direction of motion is constantly changing. Change of direction is acceleration, even at constant speed.

RideAnS [48]3 years ago

3 0

Answer:

Yes

Explanation:

A ball travelling at a constant speed in a circular structure is said to be in uniformly circular motion. In such a motion, there is a constant force (called centripetal force) that constantly "pulls" the ball towards the centre of its circular trajectory.

Since the ball experiences a force, it also experiences an acceleration, which has the same direction of the force: towards the centre of the circular trajectory.

But you may ask: why is there acceleration, if the speed is constant?

The fact is that acceleration is the rate of change in velocity of an object, not only in its speed. And velocity includes both speed and direction: therefore, an acceleration can be caused either by a change in speed, or by a change in the direction of motion. In this case of uniformly circular motion, the speed is constant, but the direction of the ball is constantly changing, so the ball is experiencing an acceleration.

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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

The most common units for expressing the density of a substance are the g/cm3.

Luden [163]

True.

Density = mass / volume, Unit = g / cm³.

This is a common unit because of its affiliation with the SI unit and because that also our popular liquid which is water = 1 g/cm³

6 0

3 years ago

A private aviation helicopter's main rotor blades rotate at approximately

Arisa [49]

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity $\omega$ a helicopter's main rotor blades:

$\omega=450 rpm=450 \frac{rev}{min}$

However, we are asked to express this $\omega$ in the International Systrm (SI) units. In this sense, the SI unit for time is second ( $s$ ):

$\omega=450 \frac{rev}{min} \frac{1 min}{60 s}$

$\omega=7.5 \frac{rev}{s}$

4 0

3 years ago

When you trace the outline of your palm how do you find its area

Tcecarenko [31]

Answer:

you count the squares or messure it

Explanation:

you can raw equal squares about 1 cm wide if possible all equal and count the squares eg theres 10 squares (small hand) so that would be 10cm squared

7 0

3 years ago

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WEATHER: <br> What kind of weather occurs along this type of front?

Radda [10]

Cold front because the warm front or hit front is lower on the right and goes downwards and cold front is in the middle and goes to Minneapolis

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2 years ago

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