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2 years ago

Give a claim about what is affecting the orbital speed and how it affects it

Physics

1 answer:

scoray [572]2 years ago

7 0

Answer:

A planet's orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun's gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun's gravitational pull, and the slower it moves in its orbit

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What are the balanced forces for someone in a parachute? A. Gravity and Earth B. Gravity and air resistance C. Gravity and centr

lapo4ka [179]

Answer:

B. Gravity and air resistance

Explanation:

Experimentation has proven that :

When the parachute opens, air resistance becomes much greater than gravity. The forces are no longer balanced and this changes the

speed

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2 years ago

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A good way to de-magnetize something is to

FromTheMoon [43]

Answer:

B. place it in a magnetic field

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2 years ago

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The heating element of a coffeemaker operates at 120 V and carries a current of 4.50 A. Assuming the water absorbs all of the en

Bond [772]

Answer:

It will take 313.376 sec to raise temperature to boiling point

Explanation:

We have given that potential difference V = 120 Volt

Current i = 4.50 A

So resistance $R=\frac{V}{i}=\frac{120}{4.50}=26.666ohm$

Heat flow in resistor will be equal to $Q=i^2Rt$

It is given that this heat is used for boiling the water

Mass of the water = 0.525 kg = 525 gram

Specific heat of water 4.186 J/gram/°C

Initial temperature is given as 23°C

Boiling temperature of water = 100°C

So change in temperature = 100-23 = 77°C

Heat required to raise the temperature of water $Q=mc\Lambda T$

So $4.50^2\times 26.666\times t=525\times 4.186\times 77$

t = 313.376 sec

So it will take 313.376 sec to raise temperature to boiling point

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2 years ago

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If a conductor wire gets heated. What happens with its resistance?

yawa3891 [41]

If a conductor wire is heated, its resistance decreases and conductivity increases.

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2 years ago

La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.

MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0

2 years ago