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Anika [276]
3 years ago
14

How long does it take venus to orbit the sun

Physics
2 answers:
ivolga24 [154]3 years ago
6 0

Venus is about 67 million miles away from the sun. It is hot there. Venus is almost the same size as planet Earth. One year on Venus is about 225 Earth days. A day on Venus is about 243 Earth days long because Venus rotates very slowly. On Venus a day is longer than a year and a year shorter than a day.

It take Venus 225 days to orbit the sun.

kodGreya [7K]3 years ago
5 0
The answer is 225 days
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Uncle Harry weighs 80 newtons. What is his mass in<br> kilograms?
Readme [11.4K]

Answer:

8.158 Kilograms Force

Explanation:

5 0
3 years ago
Read 2 more answers
A
asambeis [7]

Answer:

22Volts

Explanation:

The pd at the terminal is known as the emf

Since there are Ten 2.2V cells

Terminal voltage = number of cells * pd of one cell

Terminal voltage = 10 * 2.2

Terminal voltage = 22V

Hence the pd at the battery terminals​ is 22Volts

4 0
3 years ago
6. A child tugs on a rope attached to a 0.62-kg toy with a horizontal force
ser-zykov [4K]

Answer:

Explanation:

Force Mass * acceleration

F = ma

a = F/m

Sum of force = 16.3N - 15.8N = 0.5N

Mass = 0.62kg

Substitute

a = 0.5/ * 0.62

a = 0.81m/s²

The acceleration of the toy is 0.81m/s²

8 0
3 years ago
A vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?
GaryK [48]

Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

8a = 45 — 5

8a = 40

a = 40 / 8

a = 5m/s^2

3 0
4 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
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