Ask question

Ask question

All categories

3 years ago

5

n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011

m. How many earth days does it take for the planet ot execute one complete orbit about the sun

1 answer:

Anna35 [415]3 years ago

4 0

Answer:

F = M2 ω^2 R centripetal force of sun on planet

ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P where P is the period

P = 2 pi (M2 * R / F)^1/2

F = G M1 M2 / R^2 gravitational force on planet

P = 2 pi {R^3 / (G M1)]^1/2

P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2

P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec

1 yr = 3600 * 24 * 365 = 3.15E7 sec

P = 3.9 / 3.2 = 1.2 years

You might be interested in

WILL MARK BRAINLIEST FOR CORRCET ANSWER!!!!!!!!!

jenyasd209 [6]

Answer:

It’s C

Explanation:

8 0

3 years ago

Read 2 more answers

State the law of conversation of energy in your own words. Give an example of a situation that you have either encountered or kn

RSB [31]

That Energy Cannot Be Created Nor Destroyed

4 0

3 years ago

A gardener uses a 60-N wheelbarrow to transport two bags of fertilizer weighing W = 252-N. Determine the maximum allowable horiz

Butoxors [25]

Explanation:

Let us assume that the maximum allowable horizontal distance be represented by "d".

Therefore, torque equation about A will be as follows.

$60 \times 0.15 + 252 \times 0.15 \times 2 + 252 \times d = 2 \times 75 \times (0.7 + 0.15 + 0.15)$

d = $\frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}$

d = 0.409 m

Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.

6 0

3 years ago

Give two examples of chemical energy being transformed into electrical energy

julsineya [31]

It can be used to create electricity and heat and can be found in propane , jet fuel, gasoline and other products. Wood - Dry wood stores chemical energy . This chemical energy is released as the wood burns, and it is burns, and it is converted into heat, which is also called thermal energy, and light energy. ( hope this be helpful)

7 0

3 years ago

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

3 years ago