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GuDViN [60]
3 years ago
5

Galileo discovered that which planet undergoes phases as viewed from earth?

Physics
1 answer:
balu736 [363]3 years ago
3 0
<h2>Answer: Venus</h2>

Galileo was the first to use the telescope to observe the heavens, mainly observing the Moon, the Sun with its sunspots, Jupiter with its moons and Venus (in the early 1600s).

In the case of Venus, he observed that it presented phases (such as those of the moon) together with a variation in size; observations that are only compatible with the fact that Venus rotates around the Sun and not around Earth.

This is because Venus presented its smaller size when it is in full phase and the largest size when it is in the new one, when it is between the Sun and the Earth.

These images along with other discoveries were presented to the Catholic Church (which supported the <u>geocentric theory</u> for that time) as a proof that completely refutes Ptolemy's geocentric system and affirms <u>Copernicus' heliocentric theory.</u>

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A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot
sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

         Initial momentum = Final momentum

         4.4v = 12.364 i + 15.092 j

         v =2.81 i + 3.43 j

Magnitude

        v=\sqrt{2.81^2+3.43^2}=4.43m/s

Direction

       \theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0

       50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0
3 years ago
Explain the difference between balanced forces and action and reaction forces.
Monica [59]
Action-reaction forces<span> act on different objects; </span>balanced forces<span> act on the same object. </span>Balanced forces<span> can result in acceleration, </span>action-reaction forces<span> cannot. ... Newton's Third Law of Motion does not apply to </span>balanced forces<span>.</span>
5 0
3 years ago
Describe the difference between balanced forces and action/reaction forces
Alecsey [184]
 Balanced forces<span> act on the same object and </span>Action-Reaction forces<span> act on different objects.</span>
5 0
3 years ago
Read 2 more answers
A falcon can descend with a speed 250 km/h. If a falcon flies at this speed for 2.0 s and then flies a 100 m in 2.5 s, what is t
Vadim26 [7]

Answer:

v= s/t

Explanation:

250 km/ h =69.44m/s

S1=2 times 69.44 ≈ 139m

Next 2.5 seconds:

S2 = 100m

Average speed:

v=139m+100m/2s+2.5s = 239/4.5s = 53.2 m/s=192km/h

3 0
2 years ago
Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has
a_sh-v [17]

Answer:

  h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

         Em₀ = U = m g h

final point. Lower, just before the crash

         Em_f = K = ½ m v_{b}^2

energy is conserved

         Em₀ = Em_f

         m g h = ½ m v²

         v_b = \sqrt{2gh}

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

         p₀ = 2m 0 + m v_b

final instant. Right after the crash

         p_f = (2m + m) v

       

the moment is preserved

         p₀ = p_f

         m v_b = 3m v

         v = v_b / 3

         

          v = ⅓ \sqrt{2gh}

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

          Em₀ = K = ½ 3m v²

Final point. Higher

          Em_f = U = (3m) g h'

          Em₀ = Em_f

          ½ 3m v² = 3m g h’

           

we substitute

         h’=  \frac{v^2}{2g}

         h’ =  \frac{1}{3^2} \  \frac{ 2gh}{2g}

         h’ = 1/9 h

6 0
3 years ago
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