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bulgar [2K]
3 years ago
12

What is the difference in the positioning of the Sun, Earth and moon for a lunar eclipse versus a full moon?

Physics
1 answer:
Lady bird [3.3K]3 years ago
3 0

When the Moon passes between Sun and Earth, the lunar shadow is seen as a solar eclipse on Earth. When Earth passes directly between Sun and Moon, its shadow creates a lunar eclipse. Lunar eclipses can only happen when the Moon is opposite the Sun in the sky, a monthly occurrence we know as a full Moon.
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How do the properties of waves correspond with observations of waves?
Sveta_85 [38]

Answer:

no sabo por q no mi entender

7 0
2 years ago
A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi
Darina [25.2K]

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

5 0
3 years ago
A rock is rolling down a hill, and it’s halfway down. Would it have both Kintectic engery and Potenial engery? PLEASE HELP
dimaraw [331]

Answer:

Only kinetic.

Explanation:

Potential energy means it has the potential to move. Not something already in motion.

6 0
3 years ago
All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will
VikaD [51]

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}

since v_y = 0 at the top of its trajectory. Plugging in the numbers,

t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}

8 0
3 years ago
A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the
zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

V = \frac{\pi d^2}{4}h

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

V_0 = \frac{\pi d^2}{4}H

V_f = \frac{\pi d^2}{4}h

We know as well by definition that

1gal = 3.785*10^{-3}m^3

Then we have for the statement that

V_f = V_0 -1gal

V_f = V_0 - 3.785*10^{-3}

Replacing the previous data

\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}

\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}

Solving to get h,

h = 1.99963m

Therefore the change is

\Delta h = H-h

\Delta h = 2- 1.99963

\Delta h = 3.7*10^{-4}m=0.37mm

Therefore te change in the height of the water in the tank is 0.37mm

4 0
3 years ago
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