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2 years ago

8

The dwarf planet praamzius is estimated to have a diameter of about 300km and orbits the sun at a distance of 6.4E12m . What is

it’s orbital period in years?

1 answer:

RoseWind [281]2 years ago

8 0

Answer:

Time period of the planet is

$T_2 = 278.7 years$

Explanation:

As we know by Kepler's law of time period

square of time period of planet is proportional to the cube of its orbital radius

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

here we know for Earth

$T_1 = 1 year$

$R_1 = 1.5 \times 10^{11} m$

now we have

$\frac{1^2}{T_2^2} = \frac{(1.5 \times 10^{11})^3}{(6.4 \times 10^{12})^3}$

$T_2^2 = 7.77 \times 10^4$

$T_2 = 278.7 years$

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Hydraulic press is called an instrument for multiplication of force. Why?

Lisa [10]

Answer:

Hydraulic press is called an instrument for multiplication of force. Why? Because it uses Pascal's idea and principle: F=p*S. If we apply small force to small piston you will generate a pressure. According to Pascal's law pressure is the same everywhere in closed system so the same pressure will act on large piston on the other side too.

Explanation:

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1 year ago

Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s

PIT_PIT [208]

Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

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2 years ago

Sunlight strikes a piece of crown glass at an angle of incidence of 38.0°. Calculate the difference in the angle of refraction b

zhuklara [117]

Answer:

Difference in the angle of refraction = 0.3°

41.04° is the minimum angle of incidence.

Explanation:

Angle of incidence = 38.0°

For yellow light :

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

n₁ is the refractive index for yellow light which is 1.523

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.523}{1}$

${sin\theta_2}=0.9377$

Angle of refraction for yellow light = sin⁻¹ 0.9377 = 69.67°.

For green light :

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

n₁ is the refractive index for green light which is 1.526

n₂ is the refractive index of air which is 1

So,

$\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.526}{1}$

${sin\theta_2}=0.9395$

Angle of refraction for green light = sin⁻¹ 0.9395 = 69.97°.

The difference in the angle of refraction = 69.97° - 69.67° = 0.3°

Calculation of the critical angle for the yellow light for the total internal reflection to occur :

The formula for the critical angle is:

${sin\theta_{critical}}=\frac {n_r}{n_i}$

Where,

${\theta_{critical}}$ is the critical angle

$n_r$ is the refractive index of the refractive medium.

$n_i$ is the refractive index of the incident medium.

n₁ is the refractive index for yellow light which is 1.523 (incident medium)

n₂ is the refractive index of air which is 1 (refractive medium)

Applying in the formula as:

${sin\theta_{critical}}=\frac {1}{1.523}$

The critical angle is = sin⁻¹ 0.6566 = 41.04°

5 0

2 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

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2 years ago

A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str

Kisachek [45]

Answer:

$B_2 = 1.71 mT$

Explanation:

As we know that the magnetic field near the center of solenoid is given as

$B = \frac{\mu_0 N i}{L}$

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

$\frac{B_1}{B_2} = \frac{L_2}{L_1}$

now plug in all values in it

$\frac{2.0 mT}{B_2} = \frac{21}{18}$

$B_2 = 1.71 mT$

3 0

1 year ago