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3 years ago

What is a gas giant?

Physics

1 answer:

julia-pushkina [17]3 years ago

4 0

Answer:

A gas giant is a huge planet made of gases primarily hydrogen and helium. These gas giant planets include Jupiter, Saturn, Uranus, and Neptune.

You might be interested in

Time period from one noon to the next

matrenka [14]

That's the period of time known as one solar "day". We subdivide it into 24 slices which we call "hours". Using this system of time units, the day is about 4 minutes longer than one complete axial rotation of the Earth.

3 0

3 years ago

Calculate the net force on particle q1. First, find the direction of the force particle q2 is exerting on particle q1. Is it pus

ValentinkaMS [17]

The net force on particle particle q1 is 13.06 N towards the left.

<h3>Force on q1 due to q2</h3>

F(12) = kq₁q₂/r₂

F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)

F(12) = -14.41 N (towards left)

<h3>Force on q1 due to q3</h3>

F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)

F(13) = 1.352 N (towards right)

<h3>Net force on q1</h3>

F(net) = 1.352 N - 14.41 N

F(net) = -13.06 N

Thus, the net force on particle particle q1 is 13.06 N towards the left.

Learn more about force here: brainly.com/question/12970081

#SPJ1

8 0

1 year ago

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

3 0

2 years ago

Estimate the number of times the earth will rotate on its axis during a human’s lifetime

Fudgin [204]

Here we have to calculate the number of rotations made by earth around its own axis in the entire life time of a human being.

The average human life is 50 years .

Each year is 365.5 days

Hence $50 years =50 *365.5 days$

=18,275 days

The earth rotates around its own axis in 23 hours 56 minutes and 4 second which is approximately equal to 24 hours or one days.

Hence one rotation takes one days.

⇒18,275 days =18,275 rotations

Hence the total number of rotations made by earth around its own axis in the life time of a average human life time will be 18,275 times

7 0

3 years ago

A car with a mass of 1600 kg is towing a trailer with a mass of 420 kg. The car

Gekata [30.6K]

Answer:

1.63366

Explanation:

I got this answer from calculator soups physics calculators. I really recommend their website for formulas.

3 0

3 years ago