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Masteriza [31]
3 years ago
10

Venus's average distance from the sun is 0.72 AU and Saturn's is 9.54 AU. Calculate the orbital velocity of Venus and Saturn aro

und the sun. (The mass of the sun is 1.99 X 10^30 KG. an AU is 1.50 X 10^11 M.
Physics
1 answer:
DaniilM [7]3 years ago
4 0
So you can compute the force of gravity on the planet GMm/r^2
or (6.674 X 10^−11)(1.99 X 10^30)m / d^2(1.5 X 10^11)^2
Where d is the average distance of the planet form the sun and m is the mass of the planet (I will keep these as symbols so doing Saturn and Venus will be simple substitution into one formula in the end).

Now realize that if they are in stable circular orbit, then this force must provide the necessary centripetal force mv^2/r or mv^2 / d(1.5 X 10^11)

So we get:

(6.674 X 10^−11)(1.99 X 10^30)m / d^2(1.5 X 10^11)^2 = mv^2 / d(1.5 X 10^11)
The m's cancel out as does one 1/d(1.5 X 10^11)
(6.674 X 10^−11)(1.99 X 10^30) / d(1.5 X 10^11) = v^2
Evaluating and square rooting yeilds:
v = sqrt((8.854 X 10^8)/d)

Now plug in 0.72 for venus and 9.54 for saturn

Venus = 35067.39 m/s
Saturn = 9633.75 m/s

Some precision was lost rounding to 8.854 X 10^8, so if more presion is required just type the whole thing from before into a calculator.  Anyways the process is sound
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Answer:

Distance = 13.9 meters

Explanation:

Given the following data;

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Distance = maximum speed/decelerating speed

Distance = 41.67/3

Distance = 13.9 meters

Therefore, the bus would travel a distance of 13.9 meters before stopping.

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A 12 n cart is moving on a horizontal surface with a coefficient of kinetic friction of 0.20. what force of friction must be ove
jonny [76]

We must remember that the total net force equation at constant velocity is:

<span>F – Ff = 0</span>

of

F - µN = 0

Using Newton's 2nd Law of Motion:<span>

F = m a 

<span>Where,

F = net force acting on the body 
m = mass of the body 
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Since the cart is moving at a constant velocity, then acceleration is zero, hence the working equation simplifies to 

F = net Force = 0 

Therefore, 

F - µN = 0 

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3 years ago
What is a planet’s period of rotation?
Annette [7]


The answer is the letter "C" ( I have honors science I am good at this type of stuff )

Hope I helped :) ( ask me for help when u need it :)

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4 years ago
Read 2 more answers
A small current element carrying a current of I = 1.00 A is placed at the origin given by d → l = 4.00 m m ^ j Find the magnetic
xxTIMURxx [149]

Answer:

the magnitude and direction of d → B on the x ‑axis at x = 2.50 m is -6.4 × 10⁻¹¹T(Along z direction)

the magnitude and direction of d → B on the z ‑axis at z = 5.00 m is 1.6 × 10⁻¹¹T(Along x direction)

Explanation:

Use Biot, Savart, the magnetic field

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

Given that,

i = 1.00A

d → l = 4.00 m m ^ j

r = 2.5m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (2.5m) \hat i +(0m)^2 + (0m)^2

 =2.5m

on the axis of x at x = 2.5

r = \sqrt{(2.5)^2 + (0)^2 + (0)^2}

r = 2.5m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{2.5 \hat i}{2.5}\\\\= 1\hat i

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(2.50)^2} \\\\d\bar{B} = -6.4\times10^{-11} T

(Along z direction)

B)r = 5.00m

Displacement vector is

\bar{r}=x\hat i+y\hat j+z \hat k\\

\bar{r}= (5.00m) \hat i +(0m)^2 + (0m)^2

 =5.00m

on the axis of x at x = 5.0

r = \sqrt{(5.00)^2 + (0)^2 + (0)^2}

r = 5.00m

And unit vector

\hat r =\frac{\bar{r}}{r}

= \frac{5.00 \hat i}{5.00}\\\\= 1\hat i\\

Therefore, the magnetic field is as follow

d\bar{B}=\frac{U}{4\pi } \frac{i(d\bar{l}\times r)}{r^2}

d\bar{B} = \frac{(10^-^7)(1)(4\times10^-^3j\times i}{(5.00)^2} \\\\d\bar{B} = 1.6\times10^{-11} T

(Along x direction)

7 0
3 years ago
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