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Evgesh-ka [11]
3 years ago
14

Compared to its mass on Earth, the mass of a 60-kg object on the moon is

Physics
1 answer:
vagabundo [1.1K]3 years ago
3 0

Compared to its mass on Earth, the mass of any object on the moon is <em>precisely identical</em> to its mass on Earth, since mass doesn't depend on where the object is, or on anything else in the environment outside the object.

Its <u>WEIGHT</u>, however, on the moon is only about 16% of its weight on Earth.

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Queremos un cilindro de simple efecto que utilice en su funcionamiento un volumen de aire a presión atmosférica de 13,122 litros
zhuklara [117]

Answer:

1) El diámetro es de aproximadamente 913,987 cm.

2) La fuerza del cilindro es 5576850 kgf

Explanation:

1) Los parámetros dados son;

El volumen del aire = 13,122 litros = 13122000 cm³

La presión de trabajo = 8.5 kgf / cm²

La longitud del cilindro = 20 cm.

Por lo tanto, tenemos;

El área de la base del cilindro = π · r² = 13122000 cm³ / (20 cm) = 656100 cm²

r = √ (656100 / π) ≈ 456,994 cm

El diámetro = 2 × r ≈ 2 × 456.994 ≈ 913.987 cm

El diámetro ≈ 913,987 cm

2) La fuerza del cilindro = El área de la base del cilindro × La presión de trabajo

∴ La fuerza del cilindro = 656100 cm² × 8.5 kgf / cm² = 5576850 kgf

La fuerza del cilindro = 5576850 kgf

3 0
2 years ago
You wrap a wire around a piece of iron. If you slowly increase the strength of an electric current flowing through the wire, you
Maksim231197 [3]
The correct answer is A. the magnet to become stronger

The stronger the electric current in the piece of metal, the stronger the magnetic field will be.
4 0
3 years ago
Read 2 more answers
If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c
hichkok12 [17]

Answer:

4

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m_1 = Mass of Earth

m_2 = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

F_o=\dfrac{Gm_1m_2}{r^2}

New gravitational force

F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}

Dividing the equations

\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4

The ratio is \dfrac{F_n}{F_o}=4

The new force would be 4 times the old force

7 0
2 years ago
The latitude of the city of Arlington is about 32.7357°. Calculate the following: (a) The angular speed of the Earth. (b) The li
julia-pushkina [17]

Answer:

Explanation:

a ) The earth rotates by 2π radian in 24 x 60 x 60 s

so angular speed ( w )  = 2π / (24 x 60 x 60)  = 7.268 x  10⁻⁵ rad / s

b ) Linear speed of city of Arlington ( v )  = w r = w R Cosλ where R is radius of the earth and λ is latitude .

v = 7.268 x 10⁻⁵ x  6.371 x 10⁶ cos 32.7357

389.5 m /s

acceleration = w² r = w² R Cos 32.7357

= (7.268 x 10⁻⁵ )² x 6.371 x 10⁶ x cos 32.7357

=283.08 x 10⁻⁴ m/s²

c) velocity ratio =

w r /w R =

R cos 32.73/ R  

= Cos 32.73

= 0.84 .

6 0
3 years ago
How can a small spark start a huge explosion? using electric forces and molecules
OLga [1]

Answer:

Small sparks can lead to huge explosion if they are left unattended.

Explanation:

Small sparks are not harmful but if these sparks happen near some hazardous material or object then it could lead to heavy explosion. If there is some chemical substance near the spark or there are magnetic lines which can explode the spark then these minor sparks could result in heavy disastrous explosion.

4 0
3 years ago
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