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3 years ago

What is true about colonial families?

Physics

1 answer:

MrRissso [65]3 years ago

7 0

Answer:

Explanation:

I just took the test and C was wrong, do not listen to that answer. It is "All of the above"

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At a particular instant the magnitude of the gravitational force exerted by a planet on one of its moons is 7 × 1021 N. Collapse

9966 [12]

Answer:

Fg = 4.2*10²² N

Explanation:

The gravitational force between any two masses, provided that can be approximated by point masses (comparing their diameters with the distance between them), obeys the Newton's Universal Law of Gravitation, which states that the force (always attractive) is proportional to the product of the masses and inversely proportional to the square of the distance between them (this as a consequence of our Universe being three-dimensional), as follows:

$Fg =\frac{G*m1*m2}{r^{2}}$

So, if one of the masses increases 6 times, the force between them will be directly 6 times larger, so the new magnitude of the force will be as follows:

Fg₂ = Fg₁*6 = 7*10²¹ N* 6 = 4.2*10²² N

7 0

3 years ago

A satellite orbits the Earth in an elliptical orbit. At perigee its distance from the center of the Earth is 22500 km and its sp

aleksley [76]

Answer:

$6.09294\times 10^{24}\ kg$

Explanation:

K = Kinetic energy

$v_p$ = Perigee speed = 4280 m/s

$v_a$ = Apogee speed = 3990 m/s

$r_p$ = Perigee Distance = 22500000 m

$r_a$ = Apogee Distance = 24100000 m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Earth

m = Mass of satellite

In this system the kinetic and potential energies are conserved

$K_p+P_p=K_a+P_a\\\Rightarrow \frac{1}{2}mv_p^2-\frac{GMm}{r_p}=\frac{1}{2}mv_a^2-\frac{GMm}{r_a}\\\Rightarrow \frac{1}{2}m(v_p^2-v_a^2)+GMm\left(\frac{1}{r_a}-\frac{1}{r_p}\right)=0\\\Rightarrow M=\frac{v_a^2-v_p^2}{2G}\times \left(\frac{1}{r_a}-\frac{1}{r_p}\right)^{-1}\\\Rightarrow M=\frac{3990^2-4280^2}{2\times 6.67\times 10^{-11}}\times \left(\frac{1}{24100000}-\frac{1}{22500000}\right)^{-1}\\\Rightarrow M=6.09294\times 10^{24}\ kg$

The mass of the Earth is $6.09294\times 10^{24}\ kg$

3 0

4 years ago

Two cars are traveling around identical circular racetracks. Car A travels at a constant speed of 20 m/s. Car B starts at rest a

Tomtit [17]

Answer:

b. it has the same centripetal acceleration as car A.

Explanation:

According to the question, the data provided is as follows

Constant speed of car A = 20 m/s

Constant tangential acceleration until its speed is 40 m/s

Based on the above information, the true statement is the same centripetal acceleration as car A because

As we know that

Centripetal acceleration is

$= \frac{V^2}{r}$

where,

$V^2$ = velocity

r = radius of the path

Now if both car A and car B moving in the same or identical circular path having the same velocity so in this case there is the same centripetal acceleration for that particular time

hence, the second option is correct

3 0

4 years ago

The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic

Drupady [299]

a) Density at 100 degrees: $1.34\cdot 10^4 kg/m^3$