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3 years ago

Galileo observed that the planet __________ goes through phases like the Moon.

Physics

1 answer:

Brut [27]3 years ago

6 0

Answer:

<u>Venus</u>

Explanation:

Galileo observed in 1610 for the very first time with his first telescope that like moon Venus also exhibit different phases. Before that planets resembled bright and shining stars.

This made him excited as then he could easily register the different phase of Venus as a result of reflection of sunlight from its surface and this pave a path for him to initiate the establishment of the fact that its not earth but the sun as the center of the cosmos around which planet revolves.

When Venus was at a distance closest to our Earth earth, it seems to be bigger, but we see more of the night side of Venus which seems to be crescent-shaped.

When Venus moves away from the Earth around the Sun, it seems to be smaller, but we register more of the day side which seems to be full and round.

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a baseball pitcher throws a fastball at 98 mph. he also has a slower "changeup" pitch, 75 mph, which he throws identically, same

notsponge [240]

Answer:

For these reasons at 98 mph the path is straighter

Explanation:

To solve this problem we are going to use the kinematic equations, specifically those of projectile launches, let's calculate the distances that the ball travels

X = Vox t

Y = Yo + Voy t - ½ g t²

They tell us that the only parameter that changes is the speed, so the distance to the plate is known

t = Vox / x

We replace

Y- Yo = Voy (Vox / X) - ½ g (Vox / x)²2

Y -Yo = Vo² sinθ cos θ / x - ½ g Vo² sin²θ / x²

Y -Yo = Vo² (sinθ cosθ / x - ½ g sin²θ / x²)

The trajectory will be flatter when Y is as close as possible to Yo, when examining the right side of the equation, the amount in Parentheses is constant and to what they tell us that the angles and the distance the plate does not change.

Consequently, of the above, the only amount changes is the initial speed if it increases the square of the same increases, so that the height Y approaches the height of the shoulder, that is, DY decreases. For these reasons at 98 mph the path is straighter

7 0

3 years ago

Poorly treated municipal wastewater is discharged to a stream. The river flow rate upstream of the discharge point is Qu/s = 8.7

Musya8 [376]

Answer

given,

Q u = 8.7 m³/s

Q d= 0.9 m³/s

BOD concentration = 50 mg/L

a) BOD concentration at the down stream

$C_{down}=\dfrac{0.9\times 50}{8.7+0.9}$

$C_{down}=\dfrac{0.9\times 50}{9.6}$

= 4.69 mg/L

b) discharge = 9.6 m³/s

cross sectional area = 10 m²

velocity steam = $\dfrac{8.7+0.9}{10}$

= 0.96 m/s

time taken to move 50 km down stream = $\dfrac{50 \times 1000}{0.96}$

= 52083.3 s

= $\dfrac{52083.3}{3600\times 24}$

= 0.6 days

now,

$C_t=C_0e^{-kt}$

$C_t=4.6875\ e^{-0.2\times 0.6}$

$C_t = 4.16 mg/L$

5 0

4 years ago

A man slides on snow without friction starting at 8.96m/s at the top of an inclined plane with height 8.21m. What is his speed a

hammer [34]

Answer:

V2 = 15.53 [m/s]]

Explanation:

In order to solve this problem we must use the principle of energy conservation, where potential energy is transformed into kinetic energy. At the bottom is taken as a reference level of potential energy, where the value of this energy is equal to zero.

Above the inclined plane we have two energies, kinetics and potential. While when the sled is at the reference level all this energy will have been transformed into kinetic energy.

$E_{1}=E_{2}\\ m*g*h+(\frac{1}{2} )*m*v_{1} ^{2}=\frac{1}{2}*m*v_{2} ^{2} \\(9.81*8.21)+(0.5*8.96^{2} )=(0.5*v_{2}^{2} )\\(0.5*v_{2}^{2} )=120.68\\v_{2} ^{2}=241.36\\v_{2} =\sqrt{241.36}\\ v_{2} =15.53[m/s]$