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AysviL [449]
3 years ago
10

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

_________ year(s).
Physics
1 answer:
NARA [144]3 years ago
6 0

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

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Please help! Average speed. Show work!
KiRa [710]

Answer:

3.78 m/s

Explanation:

Recall that the formula for average speed is given by

Speed = Distance ÷ Time taken

Where,

Speed = we are asked to find this

Distance = given as 340m

Time taken = 1.5 min = 1.5 x 60 = 90 seconds

Substituting the values into the equation:

Speed = Distance ÷ Time taken

= 340 meters  ÷ 90 seconds

= 3.777777 m/s

= 3.78 m/s (round to nearest hundredth)

3 0
3 years ago
Read 2 more answers
Graph the following data tables on different graphs.
Anna [14]

Answer:

Sjjsjsjsjsjsjsjwjwjw

8 0
3 years ago
Riley is cleaning his room(a once a year activity) and pulls the Hoover with a force of 8 N a total distance of 20 metres. How m
Radda [10]

Answer:

160J

Explanation:

Given force = 8N and total distance = 20 meters

Workdone = force x distance

= 8 x 20

= 160J

Therefore, workdone by Riley in pulling the hoover is 160J

7 0
3 years ago
HELP
Free_Kalibri [48]

Answer:

Explanation:

Although there is absolutely NO regard for significant digits, I can help you with this, nonetheless.

The equation for Potential Energy is PE = mgh. We have everything but the height of the ball. We have to solve for that using a one-dimensional motion equation:

v² = v₀² + 2aΔx, where Δx is our displacement (the height we need for PE). Filling in and keeping in mind that at the max height of parabolic travel, the final velocity of the object is 0:

0 = (21.5)² + 2(-9.8)Δx and

0 = 462.25 - 19.6Δx and

-462.25 = -19.6Δx so

Δx = 23.58 m. Using this as the h in our PE equation:

PE = .19(9.8)(23.58) so

PE = 43.9 J, choice C.

8 0
3 years ago
What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?
adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 10973731.6 m⁻¹

n_f = Higher energy level = 3

n_i= Lower energy level = 2

Putting the values, in above equation, we get:

\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )

\lambda=6.56\times 10^{-7}m

Now we have to calculate the energy.

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 6.56\times 10^{-7}m

Putting the values, in this formula, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}

E=3.03\times 10^{-19}J

Therefore, the energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

3 0
3 years ago
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