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V125BC [204]
3 years ago
11

If Mercury, moved two orbital paths closer to the sun: what would be different? List serval differences, ideas.......

Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0
Well first of all, when it comes to orbits of the planets around
the sun, there's no such thing as "orbital paths", in the sense
of definite ("quantized") distances that the planets can occupy
but not in between.  That's the case with the electrons in an atom,
but a planet's orbit can be any old distance from the sun at all. 

If Mercury, or any planet, were somehow moved to an orbit closer
to the sun, then ...

-- its speed in orbit would be greater,

-- the distance around its orbit would be shorter,

-- its orbital period ("year") would be shorter,

-- the temperature everywhere on its surface would be higher,

-- if it has an atmosphere now, then its atmosphere would become
less dense, and might soon disappear entirely,

-- the intensity of x-rays, charged particles, and other forms of
solar radiation arriving at its surface would be greater.
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Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the
Aleksandr [31]

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1    ⇒   w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

           L₀ = L_f

           I₀ w₀ = I w

          w =\frac{I_o}{I} w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

          I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

        I <I₀

        I₀/I > 1    ⇒   w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

7 0
3 years ago
A spring with a constant of 16 N/m has 98 J of energy stored in it when it is extended. How far is the spring extended?
Alchen [17]

The spring has been extended for 3.5 m

<u>Explanation:</u>

We have the formula,

PE =1/2 K X²

Rewrite the equation as

PE=1/2 K d²

multiply both the sides by 2/K to simplify the equation

2/k . PE= 1/2 K  d² . 2/K

√d²=√2PE/K

Cancelling the root value and now we have,

d=√2PE/k

d=√2×98 J / 16N/m

d=√12.25

d=3.5 m

The spring has been extended for 3.5 m

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6 0
4 years ago
A satellite moves in a circular orbit at a constant speed v0 around Earth at a distance R from its center. The force exerted on
postnew [5]

Answer:

vT = v0/3

Explanation:

The gravitational force on the satellite with speed v0 at distance R is F = GMm/R². This is also equal to the centripetal force on the satellite F' = m(v0)²/R

Since F = F0 = F'

GMm/R² = m(v0)²/R

GM = (v0)²R  (1)

Also, he gravitational force on the satellite with speed vT at distance 3R is F1 = GMm/(3R)² = GMm/27R². This is also equal to the centripetal force on the satellite at 3R is F" = m(vT)²/3R

Since F1 = F'

GMm/27R² = m(vT)²/3R

GM = 27(vT)²R/3

GM = 9(vT)²R (2)

Equating (1) and (2),

(v0)²R = 9(vT)²R

dividing through by R, we have

9(vT)² = (v0)²

dividing through by 9, we have

(vT)² = (v0)²/9

taking square-root of both sides,

vT = v0/3

6 0
3 years ago
The man is walking with speed v1 = 1.35 m/s to the right when he trips over a small floor discontinuity. Estimate his angular ve
soldier1979 [14.2K]

Answer:

The angular velocity, \omega = 1.05 rad/s clockwise

Explanation:

The mass of the man, m = 72 kg

The center of mass of the man will be at the middle of his body, particularly around his abdomen.

Height of the center of mass, h = 0.71 m

The speed of the man, v₁ = 1.35 m/s

Moment of inertia about the ankle joint, I = 66 kg/m²

Based on the principle of conservation of angular momentum (about the ankle joint):

Angular momentum before impact = Angular momentum after impact

Angular momentum before impact = -mv₁h

Angular momentum before impact = -(72 * 1.35 * 0.71)

Angular momentum before impact = -69.012 kg m²/s...............(1)

Angular momentum after impact = I \omega

Angular momentum after impact = 66 * \omega.................(2)

66 \omega = -69.012\\\omega = -69.012/66\\\omega = -1.05 rad/s

The angular velocity, \omega = 1.05 rad/s clockwise

8 0
3 years ago
A cat lies on the floor. can you say that no force acts on the cat? or is it correct to say that no net force acts on the cat? e
DIA [1.3K]

Odd though it seems at first, gravity is pulling the cat down while the floor is pushing the cat up - in equal amounts. Forces are absolutely acting on the cat but they balance - so there is no net force.  


5 0
3 years ago
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