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2 years ago

What is the relationship between a planet’s distance from the sun and the length of its year? Explain.

Physics

1 answer:

laila [671]2 years ago

5 0

Answer:

Planets that are close to the sun have shorter years, and planets that are farther have longer years, because of how fast/slow it orbits the sun.

Explanation:

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A 6 V battery is connected to a 24 ohm resistor to create a circuit. The 6 V battery is then replaced with a 12 V battery. How d

11Alexandr11 [23.1K]

Answer:

1.) The current is doubled

2.) moving a magnet up and down near the wire

3.) An electric current in the wire produces a magnetic field.

4.) Distance between Particles (m)

Explanation:

1.) When 6 V battery is connected to a 24 ohm resistor to create a circuit, using ohms law, V = IR

Current I = 6/24 = 0.25 A

When the The 6 V battery is replaced with a 12 V battery,

Current I = 12/24 = 0.5 A

Therefore, The current is doubled

2.) Electric current will be induced when moving a magnet up and down near the wire

3.) An electric current in the wire produces a magnetic field.

4.) Distance between Particles (m)

The force of attraction between two different masses is inversely proportional to the square of the distance between them.

6 0

3 years ago

Why are yet planes faster tharn. cargo planes?

stellarik [79]

Answer:

cargo planes hold cargo so there hevier

Explanation:

8 0

2 years ago

A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?

Alex17521 [72]

<h2>Answer:53.63 $ms^{-2}$ </h2>

Explanation:

The equations of motion used in this question is $v=u+at$

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of $9.8ms^{-2}$ .

In X-Direction,

Given that initial velocity= $u_{x}$ = $33.8ms^{-1}$

Using $v=u+at$ ,

$v_{x}=33.8+(0)4.25=33.8ms^{-1}$

In Y-Direction,

Given that initial velocity= $u_{x}$ = $0ms^{-1}$

Using $v=u+at$ ,

$v_{y}=0+(9.8)4.25=41.65ms^{-1}$

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}$

7 0

3 years ago

Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c

stira [4]

Answer:

i) 24.5 m/s

ii) 30,656 m

iii) 89,344 m

Explanation:

Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer

i) Los parámetros dados son;

Altura inicial, s = 120 m

El tiempo en caída libre = 2.5 s

De la ecuación de caída libre, tenemos;

v = u + gt

Dónde:

u = Velocidad inicial = 0 m / s

g = Aceleración debida a la gravedad = 9.81 m / s²

t = Tiempo de caída libre = 2.5 s

Por lo tanto;

v = 0 + 9.8 × 2.5 = 24.5 m / s

ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación

s = u · t + 1/2 · g · t²

= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m

iii) La altura restante = 120 - 30.656 = 89.344 m.

6 0

3 years ago

2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?

yanalaym [24]

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

7 0

3 years ago