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kari74 [83]
3 years ago
9

Hold a pencil in front of your eye at a position where its blunt end just blocks out the Moon. Make appropriate measurements to

estimate the diameter of the Moon, given that the Earth–Moon distance is 3.8 × 105 km. Assume that the pencil has a diameter of about 0.7 cm, and it just blocks out the Moon if it is held about 0.75 m from the eye.Express your answer using two significant figures.
Physics
1 answer:
kenny6666 [7]3 years ago
4 0

Answer:

D = 3.6 \times 10^6 m

Explanation:

Since the diameter of the moon position and the diameter of pencil position are in same line

so the angle made by the rays coming from the sun will be same as the angle made by the pencil lead at eye position

so we will have

\frac{D}{L} = \frac{d}{x}

here we have

D = diameter of moon

d = diameter of pencil lead = 0.7 cm

L = 3.8 \times 10^8 m

x = 0.75 m

so we have

\frac{0.007}{0.75} = \frac{D}{3.8 \times 10^8}

D = 3.6 \times 10^6 m

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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
what is the acceleration of a softball if it has a mass of 0.50kg and hits the catcher’s glove with a force of 25 N
Kitty [74]

Answer:

mass=0.50kg

force=25N

acceleration =?

Now,

force=m×a

25=0.50×a

25÷0.50=a

50=a

acceleration =50m/s^2 answer!!!!

hope this may help you!!!!

3 0
3 years ago
Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
A substance with a define shape and volume is a
Vlad1618 [11]
Solid is the answer.
8 0
3 years ago
What is a mixture of rock and mineral fragments, organic matter, air, and water called?
Gekata [30.6K]
The answer should be soil
7 0
3 years ago
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