All categories

4 years ago

Why is the distance between the sun and Neptune dim light years

Physics

2 answers:

Bingel [31]4 years ago

4 0

.00004755 light years... Its really not far enough to measure in light years.

djyliett [7]4 years ago

3 0

The average distance from the Sun to Neptune is about 2.795 billion miles.

That's roughly 0.00048 of a light year .

You might be interested in

What is concentration

Eddi Din [679]

concentration is the ratio of solute in a solution to either solvent or total solution.

3 0

3 years ago

Read 2 more answers

The mass of the sun is 1.9891030 kg and the mass of the Earth is 5.9721024kg. If the Earth’s acceleration toward the sun is 0.00

saul85 [17]

Answer:

this is the answer according to my calculations

Explanation:

0.001.9

5 0

3 years ago

A parallel-plate capacitor is made from two aluminum-foil sheets, each 5.9 cm wide and 5.6 m long. Between the sheets is a Teflo

Zarrin [17]

Answer:

1.98 x 10⁻⁷ F

Explanation:

w = width of the sheet = 5.9 cm = 0.059 m

L = length of the sheet = 5.6 m

Area of the sheet is given as

A = L w = (5.6) (0.059) = 0.3304 m²

d = distance between the sheets = 3.1 x 10⁻⁵ m

k = dielectric constant of teflon = 2.1

Capacitance is given as

$C = \frac{k\epsilon _{o}A}{d}$

$C = \frac{(2.1)(8.85\times 10^{-12})(0.3304)}{3.1\times 10^{-5}}$

C = 1.98 x 10⁻⁷ F

5 0

4 years ago

An object of inertia 0.5kg is hung from a spring, and causes it to extend 5cm. In an elevator accelerating downward at 2 m/s^2 ,

Sidana [21]

Answer:3.98cm

Explanation:

given data

mass of object $\left ( m\right )$ =0.5kg

intial extension=5 cm

elevator acceleration=2 m/ $s^2$

From FBD of intial position

Kx=mg

K= $\frac{0.5\times 9.81}{0.05}$

k=98.1 N/m

From FBD of second situation

mg-k $x_0$ =ma

k $x_0$ =m(g-a)

$x_0$ = $\frac{0.5(9.81-2)}{98.1}$

$x_0$ =3.98cm

4 0

3 years ago

A basketball has a mass of 601 g. Moving to the right and heading downward at an angle of 29 degree to the vertical, it hits the

olga55 [171]

Answer:

Change in momentum, $\Delta p=6.3\ kg-m/s$

Explanation:

It is given that,

Mass of the basketball, m = 601 g = 0.601 kg

The basketball makes an angle of 29 degrees to the vertical, it hits the floor with a speed, v = 6 m/s

It bounces up with the same speed, again moving to the right at an angle of 29 degree to the vertical. We need to find the change in momentum. It is given by :

$\Delta p=mv\ cos\theta-(-mv\ cos\theta)$

$\Delta p=2mv\ cos\theta$

$\Delta p=2\times 0.601\times 6\times \ cos(29)$

$\Delta p=6.3\ kg-m/s$

So, the change in momentum of the basketball is 6.3 kg-m/s. Hence, this is the required solution.

6 0

4 years ago