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Ket [755]
3 years ago
14

Suppose you read in the newspaper that a new planet has been found. Its average speed in its orbit is 33 kilometers per second (

km/s). When it is closest to its star, it moves at 31 km/s, and when it is farthest from its star, it moves at 35 km/s. This story is in error because
Choose one:
A. Kepler's third law says the planet has to sweep out equal areas in equal times, so
the speed of the planet cannot change.
B. the average speed is too fast.
C. using these numbers, the square of the orbital period will not be equal to
the cube of the semimajor axis.
D. planets stay at a constant distance from their stars; they don't move closer or farther away.
E. Kepler's second law says the planet must move fastest when it is closest, not
when it is farthest away.
E. Kepler's second law says the planet must move fastest when it is closest, not
when it is farthest away.
Physics
1 answer:
Harlamova29_29 [7]3 years ago
5 0

Answer:

E. Kepler's second law says the planet must move fastest when it is closest, not  when it is farthest away.

Explanation:

We can answer this question by using Kepler's second law of planetary motion, which states that:

"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"

This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).

In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is

E. Kepler's second law says the planet must move fastest when it is closest, not  when it is farthest away.

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Neutron is commonly used to initiate a fission chain reaction.
5 0
3 years ago
Nuclear Energy
topjm [15]

Answer:

When uranium is mined, it consists of approximately 99.3% uranium-238 (U238), 0.7% uranium-235 (U235), and < 0.01% uranium-234 (U234). These are the different uranium isotopes. Isotopes of uranium contain 92 protons in the atom's center or nucleus. (The number of protons in the nucleus is what makes the atoms "uranium.") The U238 atoms contain 146 neutrons, the U235 atoms contain 143 neutrons, and the U234 atoms contain only 142 neutrons. The total number of protons plus neutrons gives the atomic mass of each isotope — that is 238, 235, or 234, respectively. On an atomic level, the size and weight of these isotopes are slightly different. This implies that with the right equipment and under the right conditions, the isotopes can be separated.

Explanation:

3 0
2 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
A light ray incident from medium 1 to medium 2, where n1&gt;n2. When the incident angle exceed the critical angle ac, the refrac
vovikov84 [41]

Explanation:

(a)

Critical angle is the angle at the angle of refraction is 90°. After the critical angle, no refraction takes place.

Using Snell's law as:

n_1\times {sin\theta_i}={n_2}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence

{\theta_r} is the angle of refraction = 90°

{n_2} is the refractive index of the refraction medium

{n_1} is the refractive index of the incidence medium

Thus,

n_1\times {sin\ \theta_{critical}}={n_2}\times{sin\ 90^0}

The formula for the calculation of critical angle is:

{sin\theta_{critical}}=\frac {n_2}{n_1}

Where,  

{\theta_{critical}} is the critical angle

(b)

No it cannot occur. It only occur when the light ray bends away from the normal which means that when it travels from denser to rarer medium.

7 0
3 years ago
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
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