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3 years ago

"The gravity of the Sun causes the planets to move in a circular path."

1 answer:

3 0

You're thinking of Kepler's first law of planetary motion, but that's not
what it says.

First of all, Kepler didn't say anything about gravity. He only described
the orbits. And he said that the orbit of each planet is an ellipse, with the
sun at one focus.

If by some chance you're thinking of Newton ... he showed that if his formula for
gravity is correct, then the orbit of a planet must be an ellipse. But Newton's law
of gravity is not one of his so-called "three laws of motion".

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Investigar la función de los espejos esféricos y como funciona, en el microscopio.

NikAS [45]

Answer:

yea

Explanation:

4 0

3 years ago

A ball is dropped from rest at a point 12 m above the ground into a smooth, frictionless chute. The ball exits the chute 2 m abo

Nonamiya [84]

Answer:

29,7 m

Explanation:

We need to devide the problem in two parts:

A) Energy

B) MRUV

Energy:

Since no friction between pint (1) and (2), then the energy conservatets:

Energy = constant ----> Ek(cinética) + Ep(potencial) = constant

Ek1 + Ep1 = Ek2 + Ep2

Ek1 = 0 ; because V1 is zero (the ball is "dropped")

Ep1 = m*g*H1

Ep2= m*g*H2

Then:

Ek2 = m*g*(H1-H2)

By definition of cinetic energy:

m*(V2)²/2 = m*g*(H1-H2) ---> $V2 = \sqrt{(2*g*(H1-H2)}$

Replaced values: V2 = 14,0 m/s

MRUV:

The decomposition of the velocity (V2), gives a for the horizontal component:

V2x = V2*cos(α)

Then the traveled distance is:

X = V2*cos(α)*t.... but what time?

The time what takes the ball hit the ground.

Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²

In the vertical axis:

Y3 = 0 ; Y2 = H2 = 2 m

Reeplacing:

-2 = 14*t + (1/2)*(-9,81)*t²

solving the ecuation, the only positive solution is:

t = 2,99 sec ≈ 3 sec

Then, for the distance:

X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m

6 0

4 years ago

Read 2 more answers

What happens to a gas when its temperature decreases?

TiliK225 [7]

Hey There!

When the gas temperature decreases the volume decreases too.

Thank You!

Have a question? Need Support? Feel free to message me.

6 0

4 years ago

Read 2 more answers

A 88.6-kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.9kg . (Use 9.80m/s2 for the gravitational acce

Dafna11 [192]

Answer:

Tension maximum =1131.9 N

Tension minimum =868.28 N

Tension at 3/4= 1065.995 N

Explanation:

Given Mass of wrecking ball M1=88.6 Kg

Mass of the chain M2=26.9 Kg

Maximum Tension Tension max=(M1+M2) × (9.8 m/s²)

=(88.6+26.9) × (9.8 m/s²)

=115.5 × 9.8 m/s²

Tension maximum =1131.9 N

Minimum Tension Tension minimum=Mass of the wrecking ball only × 9.8 m/s²

=88.6 × 9.8 m/s²

Tension minimum =868.28 N

Tension at 3/4 from the bottom of the chain =In this part you have to use 75% of the chain so you have to take 3/4 of 26.9

= (3/4 × 26.9)+88.9) × 9.8 m/s²

= (20.175+88.6) × 9.8 m/s²

=(108.775) × 9.8 m/s²

=1065.995 N

6 0

4 years ago

A record spins at 33 rpm (revolutions per minute), which is an angular velocity of about 3.46 radians per second. What is the ap

lubasha [3.4K]

Hi there!

We can use the following equation to relate angular velocity to linear velocity.

$v = \omega r$

v = linear velocity (m/s)

ω = angular velocity (3.46 rad/sec)

r = distance from axis of rotation (.12 m)

Plug in the given values.

$v = (3.46)(.12) = \boxed{.415 \frac{m}{s}}$

6 0

2 years ago