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BartSMP [9]
2 years ago
13

An extraterrestrial civilisation lives on a planet with a very elliptical orbit. Additionally, thousands of large asteroids orbi

t their solar system. The civilisation uses the light from their home star to count the number of asteroids in the direct line between the star and their planet. For a first measurement, they count the asteroids for 60 days and detect 1000 objects. Several months later, they start a second measurement: This time, they count for 80 days. How many asteroids will they detect during the second measurement? Explain why. (Note: Assume that the asteroids are homogeneously distributed in their solar system.)
i will mark as a brainiest please help me as soon ​

Physics
2 answers:
Olenka [21]2 years ago
6 0

According to Keplar's second law of planetary motion, "The areas swept by the radial vector connecting the star (sun) and the planet orbiting it are equal for equal intervals of time

Alina [70]2 years ago
5 0

For Kepler's third law this can be solved as normal ratios

  • 1000/60=x/80
  • 100/6=x/80
  • 6x=8000
  • x=8000/6
  • x=1333.34asteroids
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gregori [183]
Rutherford's experiment<span> utilized positively charged alpha particles (He with a +2 charge) which were deflected by the dense inner mass (nucleus). The conclusion that could be formed from this result was that </span>atoms<span> had an inner core which contained most of the mass of an </span>atom<span> and was positively charged.</span>
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3 years ago
A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
  • The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

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How much heat is given off when 210.0g of water at 0 degrees freezes into ice?
Sergio [31]

In the freezing physical change, when 210.0 g of water a 0 degrees freezes into ice, it gives off 71.0 kJ of heat.

<h3>What is freezing?</h3>

It is a physical change in which liquids give off heat to form solids.

We have 210.0 g of water at 0°C. We can calculate the amount of heat given off when it freezes into ice using the following expression.

Q = ΔH°fus × m

Q = 0.334 kJ/g × 210.0 g = 70.1 kJ

where,

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  • ΔH°fus is the latent heat of fusion.
  • m is the mass.

In the freezing physical change, when 210.0 g of water a 0 degrees freezes into ice, it gives off 71.0 kJ of heat.

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