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3 years ago

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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You are in Paris, 60 m up in the Eiffel Tower. If you throw a euro downward at a velocity of 2.0 m/s, how long would it take the

kondor19780726 [428]

Answer:

t = 3.29 seconds

Explanation:

It is given that,

Height of the Eiffel tower is 60 m

Initial speed of a euro, u = 2 m/s

It will move under the action of gravity in the downward direction. Firstly, we can find the final velocity as follows :

$v^2-u^2=2ad\\\\v=\sqrt{u^2+2ad} \\\\v=\sqrt{(2)^2+2\times 9.81\times 60} \\\\v=34.36\ m/s$

Let t is the time taken by the euro to hit the ground. It can be calculated as :

$v=u+at\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{34.36-2}{9.81}\\\\t=3.29\ s$

Hence, it will take 3.29 seconds to hit the ground.

4 0

3 years ago

A- what constant acceleration, in SI units, must a car have to gofrom zero to 60 mph in 10 s?

umka2103 [35]

Answer:

(a) 0.017m/s^2

(b) 17/100,000

Explanation:

(a) speed = 60mph = 60m/1h × 1h/3600s = 0.017m/s, time = 10s

Acceleration (a) = speed ÷ time = 0.017m/s ÷ 10s = 0.0017m/s^2

(b) g = 9.8m/s^2, a = 0.0017m/s^2

a/g = 0.0017/9.8 = 0.00017 = 17/100,000

Distance in foot = 0.17 × 3.2808ft = 0.558ft

3 0

3 years ago

Read 2 more answers

Hi im a little stuck on this question that came from my textbook in class it got a little confusing for me because i really dont

koban [17]

Consult the attached free body diagram.

By Newton's second law, the net force on the crate acting parallel to the surface is

∑ F[para] = (370 N) cos(-20°) - f = 0

(this is the x-component of the resultant force)

where

• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force

• f = magnitude of kinetic friction

The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.

Solve for f :

f = (370 N) cos(-20°) ≈ 347.686 N

The net force acting perpendicular to the surface is

∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0

(this is the y-component of the resultant force)

where

• n = magnitude of normal force

• 1480 N = weight of the crate

• (370 N) sin(-20°) = magnitude of the vertical component of push

The crate doesn't move up or down, so it's also in equilibrium in this direction.

Solve for n :

n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N

Then the coefficient of kinetic friction is µ such that

f = µn ⇒ µ = f/n ≈ 0.216

7 0

1 year ago

A flat, 179 179 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.41 cm 2 4.41 cm2

Alecsey [184]

Answer:

The value of the magnetic field is $B =0.1423T$

Explanation:

From the question we are told that

The number of turns is $N = 179$

The area of the loop is $A = 4.41cm^2 = \frac{4.41}{10000} = 0.000414m$

The angle is $\theta = 59^o$

The torque is $\tau =2.25 * 10^{- 5} N$

The current is $I = 2.49\ mA$

The torque acting on the current carry loop is mathematically represented as

$\tau = B * I * N * A * sin \theta$

Where is the magnitude of the magnetic filed

Making B the subject

$B= \frac{\tau}{I * N * A * sin\theta}$

Substituting values

$B = \frac{2.25*10^{-5}}{2.49*10^{-3} * 179 * 0.000414 * sin (59)}$

$=0.1423 T$

4 0

3 years ago

Olympic skier Tina Maze skis down a steep slope that descends at an angle of 30∘ below the horizontal. The coefficient of slidin

LekaFEV [45]

net force on Maze during his sliding downwards motion

$F = mgsin30 - \mu mg cos30$

now we will have

$F = mg(sin30 - 0.10cos30)$

now to find the acceleration we will have

$a = \frac{F}{m}$

$a = g(sin30 - 0.10cos30)$

$a = 9.81(0.5 - 0.10(0.866))$

$a = 4.1 m/s^2$

now to find the speed after 4 s is given by kinematics

$v_f - v_i = at$

here we know that

$v_i = 0$

$v_f - 0 = 4.1(4)$

$v_f = 16.22 m/s$

so Maze speed after 4 s will be 16.22 m/s

5 0

3 years ago