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Sergio [31]
3 years ago
6

At what position in its elliptical orbit is the speed of a planet a maximum? when it is closest to the sun when it is farthest f

rom the sun when it is midway between its farthest and closest distances to the sun everywhere in its orbit, the speed is constant (b) At what position is the speed a minimum? when it is closest to the sun when it is farthest from the sun when it is midway between its farthest and closest distances to the sun everywhere in its orbit, the speed is constant
Physics
1 answer:
Phantasy [73]3 years ago
7 0

Answer:

1.when it is closest to the sun

2.when it is midway between its farthest

Explanation:

According to the law of  Kepler's

T ² ∝ r³

T=Time period

r=semi major axis

We also know that time period T given as

T=\dfrac{2\pi r}{v}

v=Speed

v=\dfrac{2\pi r}{T}

v\alpha \dfrac{r}{T}

v\alpha \dfrac{r}{T}

v^2\alpha \dfrac{r^2}{T^2}

v^2\alpha \dfrac{r^2}{r^3}

v^2\alpha \dfrac{1}{r}

v\alpha \dfrac{1}{\sqrt{r}}

So we can say that ,when r is more then the speed will be minimum and when r is low then speed will be maximum.

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Answer:

(B) 0.5 g

Explanation:

Newton's second law says ∑  F i = m a .

the rate of change in momentum of a body is proportional to the force applied on the body.

f∝ma

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were k is constant and equal to 1

The centripetal acceleration is an acceleration.

the tension on the swing and object weight goes to the left hand side while the centripetal acceleration goes to the right handside

At the bottom of the swing, ΣF = FT – mg = mac;

notice that the tension in the swing is 1.5 times the weight of the object

we can write

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c

Explanation:

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An astronaut has a mass of 74.0 kg. 1) how much would the astronaut weigh on mars where surface gravity is 38.0% of that on eart
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A hippo drives 42 km due East. He then turns and drives 28 km at 25° East of South. He turns again and drives 32 km at 40° North
ch4aika [34]

Answer:

a) Please, see the attched figure

b) Total displacement R = (78.3 km; -4.8 km)

c) R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The hippo is 78.4 km from his starting point.

The total distance traveled is 102 km

Explanation:

a)Please, see the attached figure.

b) The vector A can be expressed as:

A = (magnitude * cos α; magnitude * sin α)

Where

magnitude = 42 km

α= 0

Then,

A = (42 km ; 0) or 42 km i

In the same way, we can proceed with the other vectors:

B = ( Bx ; By)

where

(apply trigonometry of right triangles: sen α = opposite / hypotenuse and

cos α = adjacent / hypotenuse. See the figure to determine which component of vector B is the opposite and adjacent side to α)

Bx = 28 km * sin 25 = 11.8 km

By = 28 km * cos 25 = -25.4 km (it has to be negative since it is directed towards the negative vertical region according to our reference system)

B = (11.8 km; -25.4 km) or 11.8 km i - 25.4 km j

C = (Cx; Cy)

where

Cx = 32 km * cos 40° = 24.5 km

Cy = 32 km * sin 40 = 20.6 km

C = (24.5 km; 20.6 km)

Then:

R = A+B+C = (42 km + 11.8 km + 24.5 km; 0 - 25.4 km + 20.6 km)

= (78.3 km; -4.8 km) or 78.3 km i -4.8 km j

c) R = (78.3 km; -4.8 km)

The magnitude of R is:

magnitude = \sqrt{(78.3)^{2 }+ (-4.8)^{2}}= 78.4 km

Using trigonometry, we can calculate the angle:

Knowing that

tan α = opposite / adjacent

and that

opposite = Ry = -4.8 km

adjacent = Rx = 78.3 km

Then:

tan α = -4.8 km / 78.4 km

α = -3.5°

We can now write the vector R in magnitude and direction form:

R = (78.4 km * cos (-3.5°); 78.4 km * sin (-3.5°))

d) The displacement of the hipo relative to the starting point is the magnitude of vector R calculated in c):

magnitude R = 78. 4 km

The total distance traveled is the sum of the magnitudes of each vector:

Total distance = 42 km + 28 km + 32 km = 102 km  

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The density of water under normal conditions is 1000kg / m ^ 3, so

0.429(1000) = \rho

\rho = 429kg/m^3

The density of the object is 429kg / m ^ 3

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