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kompoz
2 months ago
14

Which characteristics of Earth’s orbit are in agreement with Kepler’s second law? Check all that apply.

Physics
2 answers:
serg [3.5K]2 months ago
6 0

Explanation:  

The planet's angular velocity in an elliptical orbit changes, indicating that it moves slower when it's further from the sun and accelerates when it is nearer.

1. Yes, Earth faces the highest gravitational pull when nearest to the Sun.

2. Yes,  Earth covers the maximum distance in a thirty-day period when closest to the Sun.

3. Yes,  Earth travels with minimal tangential speed when it is at its farthest from the Sun.

4. Yes,  Earth sweeps out identical areas over equal time intervals throughout its orbit.

Kepler's second law states that the line drawn from a planet to the sun covers the same area in equal durations of time.

Maru [3.3K]2 months ago
6 0

Here is an image displaying the correct answers.

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If 2.0 mol of gas a is mixed with 1.0 mol of gas b to give a total pressure of 1.6 atm, what is the partial pressure of gas a an
serg [3582]
<span>The partial pressure of A = 1.06 atm and the partial pressure of B = 0.53 atm</span>
8 0
1 month ago
Read 2 more answers
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m2. The wave is incident u
Yuliya22 [3333]

Answer:

The total energy can be expressed as T = 169.02 \ J

Explanation:

The problem states that

The Poynting vector, which measures energy flux, equals k = 0.939 \ W/m^2

The rectangle's length is represented by l = 1.5 \ m

The width of the rectangle is w = 2.0 \ m

The duration considered is t = 1 \ minute = 60 \ s

Mathematically, the overall electromagnetic energy incident on the area is given by

T = k * A * t

where A denotes the area of the rectangle, calculated as

A= l * w

By plugging in the respective values

A= 2 * 1.5

A= 3 \ m^2

Again substituting values

T = 0.939 * 3 * 60

T = 169.02 \ J

5 0
2 months ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Maru [3345]

Answer:

The final size is nearly the same as the initial size because the increase in size1.055\times 10^{- 7} is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, m_{p} = 1.67\times 10^{- 27} kg

Initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This proton energy is \simeq 250 GeV

Thus, the speed of the proton, v\simeq c

The time to cover 1 km = 1000 m of distance is calculated as:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

According to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
1 month ago
A student wants to determine the coefficient of static friction μ between a block of wood and an adjustable inclined plane. Of t
serg [3582]

Response:

A protractor to gauge the angle between the inclined plane and the horizontal

Explanation:

The student must elevate the free end of the adjustable inclined plane until the object just begins to slide and record the angle at that precise moment. At this juncture, the frictional force is balanced by the weight component aligned with the incline. That is:

f=\mu\,* N = \mu * m g\, cos(\theta)

and  w_{//}= m\,g\,sin(\theta)

Consequently, the coefficient of static friction can be entirely established by calculating the tangent of the angle formed by the incline with the horizontal.

f = w_{//}\\\mu *\,m \,g\,cos(\theta) = m\,g\,sin(\theta)\\\mu = tan(\theta)

For this, the sole additional tool needed is a protractor for angle measurement.

7 0
1 month ago
A car with an initial velocity of 16.0 meters per second east slows uniformly to 6.0 meters per second east in 4.0 seconds. What
serg [3582]
(6-16)/4.0=-2.5 m/s²
The car's acceleration is -2.5 m/s²
5 0
2 months ago
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