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VladimirAG [237]
3 years ago
11

What is one of Kepler's laws of planetary motion?

Physics
1 answer:
iragen [17]3 years ago
4 0

Answer:

a. Planets move on elliptical orbits with the Sun at one focus.

Explanation:

Johannes Kepler was an astronomer who discovered that planets had elliptical orbits in the early 1600s (between 1609 and 1619).

The three (3) laws published by Kepler include;

I. The first law of planetary motion by Kepler states that, all the planets move in elliptical orbits around the Sun at a focus.

II. According to Kepler's second law of planetary motion, the speed of a planet is greatest when it is closest to the Sun.

Thus, the nearer (closer) a planet is to the Sun, the stronger would be the gravitational pull of the sun on the planet and consequently, the faster is the speed of the planet in terms motion.

III. The square of any planetary body's orbital period (P) is directly proportional to the cube of its orbit's semi-major axis.

Hence, one of Kepler's laws of planetary motion states that planets move on elliptical orbits with the Sun at one focus. This is his first law of planetary motion.

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What is the momentum of a 1,985-kg car going 32.5 m/s?
mojhsa [17]

Answer:

64512.5 m/kg

Explanation:

p=mv

p=1985 x 32.5

=64512.5

5 0
3 years ago
A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t
Leona [35]

Answer:

To obtain the power, we first need to find the work made by the force.

1) To calculate the work, we need the next equation:

\int\limits {F} \, dx

So the force is given by the problem so our mission is to find 'dx' in terms of 't'

2) we know that:

\frac{dV}{dt} = a = 2.6

So we have:

v = 2.6t

Then:

\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt

3) Finally, we replace everything:

\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt

After some calculation, we have as a result that the work is:

161.9638 J.

4) To calculate the power we need the next equation:

P = \frac{W}{t}

So

P = 161.9638/4.7 = 34.46 W

8 0
3 years ago
The tire above has a recommended tire pressure of 35 PSI, however, its current pressure is only 26 PSI. Which of the following c
Dima020 [189]

Answer:

option B  

Explanation:

It is given that in the tire the recommended pressure is 35 PSI however the current pressure is only 26 PSI which means that pressure in the tire is less than the recommended so the chances of blowout of the tire gets eliminated hence option A is not correct.

Having pressure less in the tire can lead to the Unstable handling of the vehicle.

so correct answer is option B  

7 0
3 years ago
A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that
gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

v(t)=At+Bt^2

Differentiating with respect to time

\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt

At t = 0

1.4=A

Hence, A = 1.4 m/s²

B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3

B = -0.10493 m/s³

At t = 5 seconds

a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2

a = 1.29507 m/s²

T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N

T = 28095.8271 N

Weight of rocket

W=2530\times 9.81=24819.9\ N

\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W

T = 1.13198 W

3 0
3 years ago
The Force Of Gravity Acting On A Childs Mass On Earth Is 490 newtons whats the childs mass
Flauer [41]
Force of gravity on an object is the weight of the object and is given by mass times accerelation due to gravity. The accerelation due to gravity is the accerelation of an object in free fall and is given by 9.8m/s^2. Given that the force of gravity acting on a child's mass on earth is 490 newtns, i.e. F = mg which means that 490 newtons = 9.8 times mass. Therefore, mass of the child is 490 / 9.8 = 50 kg.
3 0
3 years ago
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