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natulia [17]
3 years ago
6

Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca

lculate its orbital period). A. About I degree B. About 12 degrees C About 30 degrees D. About 90 degrees E. 360 degroes
Physics
1 answer:
Lana71 [14]3 years ago
6 0

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*(\sqrt{\frac{r^3}{Gm}})

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

\frac{Tsaturn}{Tearth} = \sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be \sqrt{10^3}} years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

1year * \frac{360degrees}{31.62years} = 11.38 degrees

or about 12 degrees.

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Norma-Jean [14]
Choice-D will do the job nicely.
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3 years ago
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Which statement is true for both types of transistors?
lara31 [8.8K]

For both NPN and PNP this is true:

The base is between the collector and the emitter.

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If you were in a spaceship watching a ball hover at rest (inside the spaceship) in mid-air, and the spaceship suddenly began rap
USPshnik [31]

Answer:

It will still hover until the spaceship "hits" or exerts a force on it.

Explanation:

Remember, if there is no net force, there is no acceleration or movement.

In this case, our ball is hovering in the spaceship, and in space, we can assume there is no F_g, and we can assume there is no F_N, nor no forces acting against it.

So, the ball would not move.

However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.

This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.

Think about it this way.

Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only  F_g is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.

If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.

4 0
2 years ago
Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-deg
n200080 [17]

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N

Thus, the net force over the body is:

F=(913.14N)\hat{i}+(274.87N)\hat{j}

Next, you calculate the magnitude of the force:

F=\sqrt{(913.14N)+(274.87N)^2}=953.61N

and the direction is:

\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°

7 0
3 years ago
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
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