It takes a ball 30 seconds to travel from point A to point B. What is the ball’s speed?

If an asteroid is 4 AU from the Sun, what is the period of revolution around the Sun for the asteroid?

Vedmedyk [2.9K]

Answer: 8 years

Explanation:

According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”:</em>

<em />

$T^{2}\propto a^{3}$ (1)

In other words: this law states a relation between the orbital period $T$ of a body (moon, planet, satellite, comet, asteroid) orbiting a greater body in space (the Sun, for example) with the size $a$ of its orbit.

However, if $T$ is measured in years (Earth years), and $a$ is measured in astronomical units (equivalent to the distance between the Sun and the Earth: $1AU=1.5(10)^{8}km$ ), equation (1) becomes:

$T^{2}=a^{3}$ (2)

This means that now both sides of the equation are equal.

Knowing $a=4 AU$ and isolating $T$ from (2):

$T=\sqrt{a^{3}}$ (3)

$T=\sqrt{(4 AU)^{3}}$ (4)

Finally:

$T= 8 years$ This is the period of the asteroid

3 0

3 years ago

) The angular acceleration of the disk is defined by ???? = 3t 2 + 12 rad/s, where t is in seconds. If the disk is originally ro

pshichka [43]

Answer:

a) $\theta = 52\,rad$ , b) $v_{A} = 44\cdot r \,\frac{m}{s}$ , c) $a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}$ , $a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}$

Explanation:

a) The angular motion is obtained by integrating the angular acceleration function twice:

$\alpha = 3\cdot t^{2} + 12$

$\omega = t^{3} + 12\cdot t + 12$

$\theta = \frac{1}{4}\cdot t^{4} + 6\cdot t^{2} + 12\cdot t$

The angular motion when t = 2 s. is:

$\theta = \frac{1}{4}\cdot (2\,s)^{4} + 6\cdot (2\,s)^{2} + 12\cdot (2\,s)$

$\theta = 52\,rad$

b) Let be $r$ the distance between A and the rotation axis, measured in meters. The magnitude of the angular velocity when t = 2 s. is:

$\omega = (2\,s)^{3} + 12\cdot (2\,s) + 12$

$\omega = 44\,\frac{rad}{s}$

Finally, the magnitude of the velocity is:

$v_{A} = 44\cdot r \,\frac{m}{s}$

c) The angular acceleration of the disk when t = 2 s. is:

$\alpha = 3\cdot (2\,s)^{2} + 12$

$\alpha = 24\,\frac{rad}{s^{2}}$

Lastly, the normal and tangential components at point A are, respectively:

$a_{A,n} = \omega^{2}\cdot r$

$a_{A,n} = \left(44\right)^{2}\cdot r$

$a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}$

$a_{A,t} = \alpha \cdot r$

$a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}$

6 0

3 years ago

A wave with low frequency would have relatively ________.

N76 [4]

Answer A, low energy and long wavelength.

3 0

3 years ago

Read 2 more answers

What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?

IRISSAK [1]

Answer:

The magnitude of electric field is 22.58 N/C

Solution:

Given:

Force exerted in upward direction, $\vec{F_{up}} = 3.50\times 10^{- 5} N$

Charge, Q = $- 1.55\micro C = - 1.55\times 10^{- 6} C$

Now, we know by Coulomb's law,

$F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}$

Also,

Electric field, $E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}$

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

$\vec{E} = \frac{\vec F_{up}}{Q}$

$\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}$

$\vec{E} = - 22.58 N/C$

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

$|\vec{E}| = 22.58\ N/C$

6 0

3 years ago

Which conversion is the function of a photovoltaic cell? A) sunlight to mechanical energy B) electricity to heat C) sunlight to

irakobra [83]

Answer:

C. Sunlight to electricity

Explanation:

Photo voltaic cells covert sunlight to the electricity. It absorbs the light from the sun and use that sunlight to create electric current. The number of photo voltaic cells are connected together in a solar panel to generate enough electricity to power up a home.

3 0

3 years ago