All categories

3 years ago

What is the relationship between a planet’s distance from the sun and the amount of time it takes the planet to orbit the sun?

Physics

1 answer:

S_A_V [24]3 years ago

4 0

Answer:

Kepler's third law of planetary motion

Explanation:

Kepler's third law of motion gives the relationship between a planet’s distance from the sun and the amount of time it takes the planet to orbit the sun. Its formula is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

Where

T is the time taken by the planet to orbit the sun

M is the mass of sun

G is the universal gravitational constant

a is the distance of planet form the sun

You might be interested in

Average wavelength of radio waves

Nesterboy [21]

Answer:

Radio waves have frequencies as high as 300 gigahertz(GHz)to as low as 30 hertz(Hz).At 300 GHz the corresponding wavelength is 1mm and at 30Hz is 10,000 km

7 0

3 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

4 years ago

The path of a meteor passing Earth is affected by its gravitational force and falls to Earth's surface. Another meteor of the sa

Lina20 [59]

Answer: The correct answer is option (C).

Explanation:

As it is given in the problem, the path of a meteor passing Earth is affected by its gravitational force and falls to Earth's surface. Another meteor of the same mass falls to Jupiter's surface due to its gravitational force.

According to Newton's law of universal gravitational, every particle attracts every other particles in the universe with the gravitational force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The Jupiter is the most massive planet in the solar system. It is also the largest planet in the solar system. The gravity of Jupiter on its surface is 2.4 times that of surface gravity of the Earth.

If a person weighs 100 pounds on the Earth then he would weigh 240 pounds on Jupiter.

Therefore, the correct answer is option (C), the meteor falls to Jupiter faster due to its greater gravitational force.

8 0

4 years ago

Read 2 more answers

Residential building codes typically require the use of 12-gauge copper wire (diameter 0.2053 cm) for wiring receptacles. Such c

AysviL [449]

Answer:

a) E = 4.26 W

b) E' = 6.724 W

c) copper wire is the safer option to use.

Explanation:

Given:

Diameter of the 12 gauge copper wire, d = 0.2053 cm

Thus, Radius of the 12 gauge copper wire, r = 0.2053 cm

/ 2 = 0.10265 cm = 0.10265 × 10⁻² m

Now.

the area (A) comes out as

A = π × (0.10265 × 10⁻²)²

A = 3.3103 × 10⁻⁶ m²

Length of the copper wire, L = 2.10 m

a) The resisitivity (ρ) of copper = 1.68 × 10⁻⁸ ohm m

Now,

the resistance of the copper , R = ρL/A

R = (1.68 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

R = 0.01065 ohm

The Energy (E) is given as,

E = I²R

where, I is the current

I = 20.0 A

on substituting the values, we get

E = 20.0² × 0.01065

E = 4.26 W

(b) For the aluminium

Resisitivity, ρ' = 2.65 × 10⁻⁸ ohm m

Now, the resistance of the aluminium wire, R' = (ρ' × L) / A

Since the cross-section of the aluminium wire is same as the copper wire

thus,

R = (2.65 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

R = 0.0168 ohm

Therefore,

The Rate of energy produced by the aluminium wire, E' = I²R'

E' = 20.0² × 0.0168

E' = 6.724 W

(c) From the above results, we can conclude that the power consumed or the rate of energy produced by the aluminium wire is more.

Hence, copper wire is the safer option to use.

8 0

4 years ago

A 1.0-kg block of aluminum is at a temperature of 50°C. How much thermal energy will it lose when its temperature is reduced by

Marat540 [252]

Answer:

22425 J

Explanation:

From the question,

Applying

Q = cm(t₂-t₁).................. Equation 1

Where Q = Thermal Energy, c = specific heat capacity of aluminium, m = mass of aluminium, t₂ = Final Temperature, t₁ = Initial Temperature.

Given: c = 897 J/kg.K, m = 1.0 kg, t₁ = 50 °C, t₂ = 25 °C (The final temperature is reduced by half)

Substitute these values into equation 1

Q = 897×1×(25-50)

Q = 897×(-25)

Q = -22425 J

Hence the thermal energy lost by the aluminium is 22425 J

5 0

3 years ago