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alexira [117]
3 years ago
12

Which characteristic of the gas giants decreases with increasing distance from the sun?

Physics
2 answers:
hodyreva [135]3 years ago
7 0
Equator diameter and Number of moons. So both a and c would be the answer :) 
anyanavicka [17]3 years ago
4 0
Gas giant are plants with mass 10 times bigger than the mass of the Earth. They are also called outer planets. There are four gas giant planets in our solar system: Jupiter, Saturn, Uranus and Neptune. The farther away from the Sun they are (the distance from the Sun is increased) the longer they orbit the Sun because of the great distance. 
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Umnica [9.8K]
Since the baseball is half the weight of the softball, the baseball has to go twice as fast as the softball.

So the best answer would be D
5 0
3 years ago
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MOST elements on the periodic table are
lidiya [134]

Answer:

B

Explanation:

6 0
3 years ago
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A 20.0 kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at 34.0° above horizontal. A constant horizont
ankoles [38]

Answer:

987 joules, 3.01s

Explanation:

(A)

from the attached diagram

net force, Fnet, pulling the crate up the ramp is given by

Fnet = FcosФ - WsinФ - Fr

where FcosФ is the component of horizontal force 290N resolved parallel to the plane

WsinФ = mgsinФ = component of the weight of the crate resolved parallel to the plane

Fr = constant opposing frictional force

Fnet = 290cos34⁰ - 20 × 9.8 × sin34° - 65

Fnet = 240.421 - 109.602 - 65

Fnet = 65.82N

Work done on the crate up the ramp, W, is given by

W = Fnet × d (distance up the plane)

W = 65.819 × 15

W = 987.285 joules

W = 987 joules (to 3 significant Figures)

(B)

to calculate the time of travel up the ramp

we use the equation of motion

s = ut + \frac{1}{2}at^{2}

where s = distance up the plane, 15m

u = Initial velocity of the crate, which is 0 for a body that is initially at rest

a = acceleration up the plane, given by

a = \frac{Fnet}{m}

where m = mass of the crate, 20 kg

now, a = \frac{65.819}{20} \\a = 3.291\frac{m^{2} }{s}

from, s = ut + \frac{1}{2}at^{2}

15 = 0*t + \frac{1}{2}* 3.291 * t^{2}

15 = 0 + 1.645t^{2}

15 = 1.645t^{2}

t = \sqrt{\frac{15}{1.645} }

t = 3.019

t = 3.01s (to 3 sig fig)

7 0
3 years ago
A ray of yellow light (f = 5.09 × 1014hz) travels at a speed of 2.04 × 108meters per second in
denis23 [38]
Velocity = fλ

where f is frequency in Hz, and λ is wavelength in meters.

2.04 * 10⁸ m/s =  5.09 * 10¹⁴  Hz   *  λ

(2.04 * 10⁸ m/s) / (5.09 * 10¹⁴  Hz ) = λ

4.007*10⁻⁷  m =  λ

The wavelength of the yellow light = 4.007*10⁻⁷  m
8 0
3 years ago
An uncharged metal sphere, A, is on an insulating base. A second sphere, B, of the same size, shape, and material carrying charg
solong [7]

Answer:

0

Explanation:

  • If we bring the charged sphere B close to, but not touching it , to the uncharged sphere A, as charges can move freely on the conductor, a charge -Q will be built on the outer surface of the sphere A, facing to sphere B.
  • As the sphere A must remain neutral, a charge Q will be built on the surface, on the side farther to the sphere B, as the following condition must be met:

       Q +(-Q) =0.

  • If we now remove sphere B, and place it far away, there will be a charge redistribution within sphere A, making to disappear the separation between Q and -Q.
  • The total charge on sphere A must be 0, as there is no charge transfer from sphere B to sphere A.
3 0
3 years ago
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