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nikitadnepr [17]
3 years ago
6

PLEASE HELP 100 POINTS! Please fill in the scale distance from sun and diversion factor, listing off the numbers works

Physics
1 answer:
steposvetlana [31]3 years ago
7 0

Solved your another question same like this with scaling to Cm this time we go with metre(m)

Scale factor

  • 1km=10³m
  • 1m=10^{-3}km

Mercury

  • 58000m

Ven us

  • 108000m

Earth

  • 150000m

Mars

  • 228000m

Jupiter

  • 778000m

Saturn

  • 1430000m

Uranus

  • 2870000m

Neptune

  • 4500000m

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Air is enclosed in a syringe at room temperature. The end is the pushed so the air only fills a 1/3 of the length of the syringe
borishaifa [10]

Answer:

A) Does not change

B) Decrease

C) Increasing

D) Increasing

Explanation:

When it pushes it's volume is reduced by 1/3 because it's Area is constant there for only volume is decreased. Then the number of molecules don't change

A) don,t change

B) Decreases

C) Density is increased because it mass doesn't change. there for

Mass = Density * Volume

When Volume is decreasing to keep mass constant Density will be increased.

D) Pressure is increased according to the boil's law. it says that for fixed mass and fixed temperature pressure is inversely proportional to the volume. then volume decrease pressure increase  

4 0
3 years ago
Read 2 more answers
2. ¿Puede haber trabajo en un sistema si no hay movimiento
Andru [333]

Answer:

d. No, porque la ecuación de trabajo lo define.

Explanation:

En Física, el trabajo realizado se puede definir como la cantidad de energía transferida cuando un objeto o cuerpo se mueve a lo largo de una distancia debido a la acción de una fuerza externa.

Matemáticamente, el trabajo realizado viene dado por la fórmula;

W = F * d

<u>Dónde;</u>

  • W es el trabajo realizado.
  • F representa la fuerza que actúa sobre un cuerpo.
  • d representa la distancia recorrida por el cuerpo.

Por lo tanto, podemos deducir de la definición de trabajo y su fórmula que el trabajo se realiza cuando un objeto (cuerpo) se mueve una distancia o experimenta cualquier forma de desplazamiento mientras transfiere energía.

7 0
3 years ago
In a physics laboratory experiment, a coil with 250 turns enclosing an area of 10.5 cm2 is rotated during the time interval 3.10
olchik [2.2K]
I guess the problem is asking for the induced emf in the coil.

Faraday-Neumann-Lenz states that the induced emf in a coil is given by:
\epsilon = -N \frac{\Delta \Phi}{\Delta t}
where
N is the number of turns in the coil
\Delta \Phi is the variation of magnetic flux through the coil
\Delta t is the time interval

The coil is initially perpendicular to the Earth's magnetic field, so the initial flux through it is given by the product between the magnetic field strength and the area of the coil:
\Phi_i = BA=(5.30 \cdot 10^{-5}T)(10.5 \cdot 10^{-4} m^2)=5.57 \cdot 10^{-8} Wb
At the end of the time interval, the coil is parallel to the field, so the final flux is zero:
\Phi_f = 0

Therefore, we can calculate now the induced emf by using the first formula:
\epsilon = -N  \frac{\Delta \Phi}{\Delta t}=- (250)  \frac{5.57 \cdot 10^{-8} Wb - 0}{3.10 \cdot 10^{-2} s} = -4.5 \cdot 10^{-4} V
7 0
4 years ago
The index of refraction for red light in water is 1.331 and that for blue light is 1.340. A ray of white light enters the water
Alex Ar [27]

Answer:

(a) 47.08°

(b) 47.50°

Explanation:

Angle of incidence  = 78.9°

<u>For blue light : </u>

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₂ is the refractive index for blue light which is 1.340

n₁ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{78.9}^0}=\frac {1}{1.340}

{sin\theta_2}=0.7323

Angle of refraction for blue light = sin⁻¹ 0.7323 = 47.08°.

<u>For red light : </u>

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₂ is the refractive index for red light which is 1.331

n₁ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{78.9}^0}=\frac {1}{1.331}

{sin\theta_2}=0.7373

Angle of refraction for red light = sin⁻¹ 0.7373 = 47.50°.

5 0
4 years ago
When a small object is launched from the surface of a fictitious planet with a speed of 52.9 m/s, its final speed when it is ver
NISA [10]

Answer:

The escape speed of the planet is 41.29 m/s.

Explanation:

Given that,

Speed = 52.9 m/s

Final speed = 32.3 m/s

We need to calculate the launched with excess kinetic energy

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

K.E=\dfrac{1}{2}\times m\times(32.3)^2

We need to calculate the escape speed of the planet

Using formula of kinetic energy

\text{escape kinetic energy}=\text{launch kinetic energy}-\text{excess kinetic energy}

\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2

\dfrac{1}{2}\times v^2=\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2

v=\sqrt{2\times(\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2)}

v=41.29\ m/s

Hence, The escape speed of the planet is 41.29 m/s.

4 0
3 years ago
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