Does groundwater flow or stay still?

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

vodka [1.7K]

We will use formula for the orbital velocity of Venus, which is v = 35.02 km/s.
An average distance to the Sun ( In kilometers ) is:
R = 0.723 * 149,579,871 km= 108,150,260 km.
Than we will calculate the orbital period ( T ).
v = 2 π R / T
T = 2 π R / v
T = 2 * 3.14 * 108,150,260 km / 126,072 km/s
T = 5389.75 s ≈ <span>224.5 days
The orbital period of Venus is approximately 224.5 days.</span>

6 0

4 years ago

A runner is jogging in a straight line at a steady Vr= 3.4km/hr. When the runner is L=8km from the finish line, a bird begins fl

Lynna [10]

Answer:

The cumulative distance that the bird travel is 13.33 Km

Explanation:

Given that,

Velocity of runner = 3.4 km/hr

Distance = 8 km

Velocity of bird = 17 km/hr

Let x is the distance from the origin where the runner run into the bird

We need to calculate the value of x

Using time of runner and bird

$t_{r}=t_{b}$

$\dfrac{d_{r}}{v_{r}}=\dfrac{d_{b}}{v_{b}}$

Put the value into the formula

$\dfrac{x}{3.4}=\dfrac{8+(8-x)}{17}$

$17x=3.4\times8+3.4(8-x)$

$17x=27.2+27.2-3.4x$

$17x+3.4x=54.4$

$x=\dfrac{54.4}{20.4}$

$x=2.67$

We need to calculate the cumulative distance that the bird travel

Using distance of bird

$d_{b}=16-x$

Put the value into the formula

$d_{b}=16-2.67$

$d_{b}=13.33\ km$

Hence, The cumulative distance that the bird travel is 13.33 Km

7 0

3 years ago

A ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum he

Andrew [12]

Answer:

10.2 metres

Explanation:

Given that a ball is projected at an initial speed of 20.0 meter per second making an angle of 45.0 with horizontal. What is the maximum height it will reach?

Solution

To get the maximum height, let us use the formula

V^2 = U^2 sin^2ø - 2gH

At maximum height V = 0

U^2 sin^2ø = 2gH

Substitute all the parameters into the formula

20^2 ( sin 45 )^2 = 2 × 9.8 × H

400 × 0.5 = 19.6 H

Make H the subject of formula

H = 200 / 19.6

H = 10.204 metres.

Therefore, the maximum height reached by the projected ball is 10.2 metres.