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2 years ago

6. Mithra is an unknown planet that has two airless moons. A and B, in circular orbits around it.

Physics

1 answer:

Ainat [17]2 years ago

4 0

The acceleration of gravity on Moon B is D) 0.25 m/s2

Explanation:

The data listed in the problem are not clear: find them in the table attached.

The acceleration of gravity on a planet is given by the following equation:

$g=\frac{GM}{R^2}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

M is the mass of the planet

R is the radius of the planet

Here we want to find the acceleration of gravity on the surface of Moon B, which has the following data:

$M=1.5\cdot 10^{20} kg$ (mass)

$R=2.0\cdot 10^5 m$ (radius)

And substituting into the equation, we find:

$g=\frac{(6.67\cdot 10^{-11})(1.5\cdot 10^{20})}{(2.0\cdot 10^5)^2}=0.25 m/s^2$

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Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th

yuradex [85]

Answer:

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

So v=E X d

V=4x8=32V

k.e=eV= 2X32=64eV

3 0

2 years ago

A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th

Tju [1.3M]

Answer:

$a = \frac{2}{3}g$

$T = \frac{mg}{3}$

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

$mg - T = ma$

also by torque equation we can say

$TR = I\alpha$

$TR = \frac{1}{2}mR^2(\frac{a}{R})$

$T = \frac{1}{2}ma$

Now we have

$mg - \frac{1}{2}ma = ma$

$mg = \frac{3}{2}ma$

$a = \frac{2}{3}g$

Also from above equation the tension force in the string is

$T = \frac{1}{2}ma$

$T = \frac{mg}{3}$

7 0

2 years ago

This is really urgent

hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is $n=\frac{c}{v}$

26) Critical angle: $[tex]sin \theta_c = \frac{n_2}{n_1}$ [/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

with

$n_1, n_2$ are the refractive index of 1st and 2nd medium

$\theta_1, \theta_2$ are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

A material with refractive index of 1.1
A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$ (1)

For light moving from air to water:

$n_1 \sim 1.00$ is the index of refraction of air

$n_2 = 1.33$ is the index of refraction ofwater

In this case, $\frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75$

For light moving from air to glass,

$n_2 = 1.51$ is the index of refraction of glass

And so

$\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66$

From eq.(1), we see that the angle of refraction $\theta_2$ is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

$n=\frac{c}{v}$

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

The higher the index of refraction, the slower the light
The lower the index of refraction, the faster the light

26)

From Snell's law,

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

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, $n_1 > n_2$ , so $\frac{n_1}{n_2}>1$ , and since $sin \theta_2$ cannot be larger than 1, there exists a maximum value of the angle of incidence $\theta_c$ (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

$sin \theta_c = \frac{n_2}{n_1}$

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

$sin \theta_c = \frac{n_2}{n_1}$

For the interface glass-air,

$n_1 \sim 1.51\\n_2 = 1.00$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}$

For the interface glass-water,

$n_1 \sim 1.51\\n_2 = 1.33$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}$

So, the critical angle is larger for the glass-water interface.

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7 0

2 years ago

A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy

SashulF [63]

Answer:

$5.444\times 10^{-4}$

Explanation:

The momentum of the neutron before and after the decay is the same since there's no external force.

$P_{sys}=const\\\\P=mv\\\\K=0.5mv^2$

#The neutron is initially at rest, so after the decay:

$P_A+P_B=0\\\\P_A=-P_B$

#After decay, the proton has +ve direction with a velocity $v_A$ while the electron moves in a negative direction with a velocity $v_B$

Therefore:

$P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B$

Let the energy released during the decay be Q:

$Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}$

Hence,Kp/Ktot is 5.444x10^(-4)

4 1

2 years ago

Which of the following statements best represents the impact of evolutionary theory on the field of psychology?

grandymaker [24]

Answer- the correct option for the question would be choice number c which tells us about the natural selection process of human species.

Explanation- Evolution of psychology was studied under the given heads namely Behaviorism cognitive functionalism sociocultural Psychodynamic and structuralism.

On studying the facts under these prospects the evolution of psychology gives us the idea of the selection by nature of individuals and the pattern of their behavior shown in different aspects.

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2 years ago