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Airida [17]
2 years ago
5

Venus has an average distance to the sun of 0.723 AU. In two or more complete sentences, explain how to calculate the orbital pe

riod of Venus, and then calculate it.
Physics
1 answer:
Otrada [13]2 years ago
5 0
For this task we need to use Kepler's third law:

T = 2*pi*\sqrt{ \frac{ r^{3}}{G*M} }

where T is orbital period in seconds, r is Venus's semi-major axis, G is gravitational constant and M is mass of the sun.

Distance from earth to sun is 1AU so if we know earths period and distance from the sun we can calculate Venus period.
Te-earths period
Tv-venus period

\frac{Te}{Tv} =  \sqrt{ \frac{ Re^{3} }{ Rv^{3} } }

From the text we know that Re/Rv = 1/0.723

Which means that:
Tv =  \sqrt{0.723^3} * Te

Tv = 0.614 Te
You might be interested in
A woman is applying 300N/m2 of pressure on to door with her hand. Her hand has area of 0.02m2. Work out the force being applied​
never [62]

Answer:

6N

Explanation:

Given parameters:

Pressure applied by the woman  = 300N/m²

Area = 0.02m²

Unknown:

Force applied  = ?

Solution:

Pressure is the force per unit area on a body

        Pressure  = \frac{force}{area}

         Force  = Pressure x area

        Force  = 300 x 0.02  = 6N

8 0
2 years ago
89. An electron is moving in a straight line with a velocity of 4.0×105 m/s. It enters a region 5.0 cm long where it undergoes a
ddd [48]

Explanation:

Given that,

Initial speed of the electron, u=4\times 10^5\ m/s

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, a=6\times 10^{12}\ m/s^2  

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

v^2=u^2+2as

v^2=(4\times 10^5)^2+2\times 6\times 10^{12}\times 0.05

v = 871779.788 m/s

or

v=8.71\times 10^5\ m/s

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

t=\dfrac{v-u}{a}

t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}

t=7.85\times 10^{-8}\ s

Hence, this is the required solution.

4 0
2 years ago
Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^
ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

T \approx e^{-2CL}

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)

Next, you divide the equation (3) into (4), and finally, you solve for L':

\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0
2 years ago
Two identical carts travel at the same speed toward each other, and then a collision occurs. The graphs show the momentum of eac
madam [21]

Explanation :

The interaction between two objects is termed as the collision. The collision can be of two types i.e. elastic collision and inelastic collision.

In this case, two identical carts travel at the same speed toward each other, and then a collision occurs. In an inelastic collision, the momentum before and after the collision remains the same but its kinetic energy gets lost.

After the collision, both the object sticks over each other and moves with one velocity.

Out of the given graph, the graph that shows a perfectly inelastic collision is attached. It shows that after the collision both the carts move with the same velocity.

5 0
2 years ago
Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
kodGreya [7K]

Answer:

1.6675×10^-16N

Explanation:

The force of gravity that the space shuttle experiences is expressed as;

g = GM/r²

G is the gravitational constant

M is the mass = 1.0 x 10^5 kg

r is the altitude = 200km = 200,000m

Substitute into the formula

g = 6.67×10^-11 × 1.0×10^5/(2×10^5)²

g = 6.67×10^-6/4×10^10

g = 1.6675×10^{-6-10}

g = 1.6675×10^-16N

Hence the force of gravity experienced by the shuttle is 1.6675×10^-16N

7 0
2 years ago
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