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Elan Coil [88]
3 years ago
15

An object has a mass of 1kg on Earth. What is its weight on the moon?

Physics
1 answer:
enyata [817]3 years ago
5 0

Answer:

the Mass of 1 kg object is same in Earth & Moon.

Explanation:

Weight, on the otherhand does change with location depends on the gravity. so the answer is : Weight of one kilo on the surface of moon is 1.622 N. A body is taken from the center of the Earth to the Moon.

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If the voltage across a circuit element of constant resistance is doubled, how is the current through the circuit element affect
Karo-lina-s [1.5K]

Answer:

The current through the circuit element will also be doubled.

Explanation:

The relationship between voltage and current of a circuit element is given by Ohm's Law. According to Ohm's Law:

Voltage = (Resistance)(Current)

If the resistance of the element is kept constant, the relation between Voltage and Current through that element becomes as follows:

Voltage = (Constant)(Current)

Voltage α Current

Thus, the voltage is directly proportional to the current for constant value of the resistance.

Therefore, when the voltage across a circuit element of constant resistance is doubled, <u>the current through the circuit element will also be doubled.</u>

5 0
3 years ago
Una onda sonora se produce durante 1,5 s. Posee una longitud de onda de 2,4 m y una velocidad de 340 m/s.
cestrela7 [59]

Answer:

Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia de potencial de 120 V, durante 18 minutos ¿Qué cantidad de calor produce?, expresado en calorías

Explanation:

Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia de potencial de 120 V, durante 18 minutos ¿Qué cantidad de calor produce?, expresado en calorías

3 0
3 years ago
If two stars have the same absolute magnitude, what can be a reason for the difference in their brightness?
AysviL [449]

B. their distances from the sun.

Explanation:

Absolute Magnitude:

Astronomers defines the absolute magnitude of a stars brightness in terms of how bright a star appears from a standard distance of 10 parsecs. Parsec is a unit of distance in astronomy. 10 parsecs is equal to 32.6 light years.

Apparent Magnitude:

Apparent magnitude of a star refers to how bright the star appears at its distance from the Earth.

If two stars have the same absolute magnitude but their apparent magnitude differs, the reason is that the distance of both the stars from the Earth varies. Hence their brightness differs when measured from Earth. The farther a star is from the Earth, the fainter its brightness.

Keywords: star, brightness, parsec, light years, apparent magnitude, absolute magnitude

Learn more about stars and absolute magnitude from:

brainly.com/question/13002384

brainly.com/question/1384449

#learnwithBrainly

5 0
3 years ago
What is the resultant displacement? Explain please to
frutty [35]
The resultant<span> is the vector sum of 2 or more vectors. It is the conclusion of adding 2 or more vectors together. If </span>displacement <span>vectors A, B, and C are added together, the result will be vector R.</span>
8 0
3 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t
nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

\Phi = BA Cos \theta

Here,

\theta = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

\epsilon= -N \frac{d\Phi }{dt}

\epsilon = -N \frac{(BAcos\theta)}{dt}

\epsilon = -NAcos\theta \frac{dB}{dt}

\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)

\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )

At time t = 5.71s,  Induced emf is,

\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)(  (3.05T/s)-(13.9T/s)(5.71s))

\epsilon = 10.9V

Therefore the magnitude of the induced emf is 10.9V

4 0
3 years ago
Read 2 more answers
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