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3 years ago

Someone please help me

Physics

2 answers:

Murrr4er [49]3 years ago

6 0

Answer:

A: any mirror.

hope it's help you

Serhud [2]3 years ago

4 0

A mirror will do this

You might be interested in

What is Brewster's angle for an air-glass n = 1.58 surface

arlik [135]

n₂ = index of refraction of glass = 1.58

n₁ = index of refraction of air = 1

θ = Brewster's angle

Brewster's angle is that incident angle at which the the reflected light from the interface of two medium is completely polarized.

Brewster's angle is given as

θ = tan⁻¹(n₂ /n₁ )

θ = tan⁻¹(1.58/1 )

θ = 57.7 deg

hence the Brewster's angle comes out to be 57.7 deg

4 0

3 years ago

Suppose you had 10 identical molecules enclosed by a box. At a given instant, one molecule has an energy of 100 Joules, and the

boyakko [2]

Answer:

A) K_average = 1/20 m v², B) low entropy , C) entropy increases.

Explanation:

A) The kinetic energy of each molecule

K = ½ m v²

The average kinetic energy is the sum of each kinetic energy among the number of them

K_average = (½ m v² + 0 + 0 +) 10

K_average = 1/20 m v²

B) Entropy is the sum of the states of each molecule, in this configuration there are only two states one with energy and the other with zero energy, so it is a system with low entropy.

S = k ln W

Where S is the entropy, k the Bolztmann constant, W the amount of state present in the system in this case is 2

C) Let's start by analyzing the entropy, as time goes by the molecule that is moving collides with the other molecules and transfers them some energy, so the other molecules move to a different state, after a little In time all the molecules will have the same energy, each one in a different state or volume, so the number of possible state increases to 10, so the entropy increases.

Now let's analyze what happens with the energy of the system, for this case we have two possibilities

- The system is isolated, therefore as it cannot exchange energy with the environment, the total energy remains constant even when the energy of each molecule can fluctuate.

- If the system is not isolated, it can exchange energy with the environment, therefore the total energy changes, depending on the difference in energy between the molecules and the environment

4 0

4 years ago

A comet is approaching Venus on a parabolic path with perigee distance of 18,200 km . Calculate the total time of travel (in hou

Anna35 [415]

Answer:

<em>The time traveled is 1.39 hrs</em>

Explanation:

Equation of Trajectory of a comet is given as

$r=\frac{h^2}{\mu}\frac{1}{1+cos \theta}$

Here

h is the specific angular momentum given as

$h=v_p r_p$

μ is gravitational parameter whose value is $3.24859 \times 10^{14} \, \, m^3/s^2$ for Venus
r_p is the perigee distance of parabolic which is 18200 km
As the path of comet is parabolic so energy is conserved i.e

$\frac{1}{2}m_c v_p^2-\frac{\mu m_c}{r_p}=0\\v_p=\sqrt{\frac{2 \mu}{r_p}}$

So h is given as

$h=v_pr_p\\\\h=\sqrt{2 \mu r_p}\\h=\sqrt{2 \times 3.24859 \times 10^{14} \times 18200 \times 10^3}\\h=1.087 \times 10^{11}$

So for point a where r=24500 km

$r_1=\frac{h^2}{\mu}\frac{1}{1+cos \theta_1}\\24500 \times 10^3=\frac{1.1824 \times 10^{22}}{3.24859 \times 10^{14}}\frac{1}{1+cos \theta_1}\\0.6731=\frac{1}{1+cos \theta_1}\\1+cos \theta_1=\frac{1}{0.6731}\\cos \theta_1=1.4857-1\\cos \theta_1=0.4857\\\theta_1=cos^{-1}0.4857\\\theta_1=1.0636 rad$

So for point a where r=39000 km

$r_2=\frac{h^2}{\mu}\frac{1}{1+cos \theta_2}\\39000 \times 10^3=\frac{1.1824 \times 10^{22}}{3.24859 \times 10^{14}}\frac{1}{1+cos \theta_2}\\1.0715=\frac{1}{1+cos \theta_2}\\1+cos \theta_2=\frac{1}{1.0715}\\cos \theta_2=0.9332-1\\cos \theta_2=-0.0667\\\theta_2=cos^{-1}(-0.0667)\\\theta_2=1.6375 rad$

So as per the Barkers equation

$t_1-T=\sqrt{\frac{2 r_p^3}{\mu}}(D_1+\frac{D_1^3}{3})$

where

$D_1=tan (\theta_1/2)\\D_1=tan(0.5318)\\D_1=0.5883$

$t_2-T=\sqrt{\frac{2 r_p^3}{\mu}}(D_2+\frac{D_2^3}{3})$

where

$D_2=tan (\theta_2/2)\\D_1=tan(0.8187)\\D_2=1.0690$

$t_2-t_1=\sqrt{\frac{2 r_p^3}{\mu}}(D_2+\frac{D_2^3}{3}-D_1+\frac{D_1^3}{3})\\t_2-t_1=\sqrt{\frac{2 (18200 \times 10^3)^3}{3.24859 \times 10^{14}}}((1.0690)+\frac{(1.0690)^3}{3}-(0.5883)-\frac{(0.5883)^3}{3})\\t_2-t_1=\sqrt{\frac{2 (18200 \times 10^3)^3}{3.24859 \times 10^{14}}}(0.8201)\\t_2-t_1=(6092)(0.8201)\\t_2-t_1=4996.21 s\\t_2-t_1=1.39 hrs\\$

So the time traveled is 1.39 hrs

3 0

3 years ago

Two boys are playing with two different balls of masses m and 2 meter respectively. if first boy through vertically up and the s

aalyn [17]

The ratio of the height attended by the object thrown upward to that thrown at some angle θ with the vertical is 1.

Note: It is assumed that the mass of the first object is m and the mass of the second object is 2m. Also, the first object is thrown upward and the second object is thrown at some angle θ with the vertical.

Projectile motion: Any object that is thrown in the air is called a projectile and the motion described by it under gravity is called the projectile motion.

If the air resistance is neglected, then the acceleration due to gravity g is the same for all objects irrespective of their masses. It is given that both objects remain in the air for the same time period. So first calculate the time period for the objects when they are in the air.

Time period of the first object: The first object is thrown in the air upwards so from the second kinematics equation,

h1=u1*t1-(1/2)*gt1^2

where h1 is the height, t1 is the time, and u1 is the initial velocity for the first object.

When the object is not in air h1=0, so

0=u1*t1-(1/2)*gt1^2

After solving the above quadratic equation, the values of t1 obtained are t1=0 which represents the initial time, and t1=2u1/g which represents the time period. So the time period of the first object is,

t1=2u1/g

Time period of the second object: The second object is thrown at some angle θ with the vertical as shown in the diagram. From the diagram, the initial velocity along the vertical direction is,

u2=uocos(θ)

where uo is the initial velocity and u2 is the initial velocity along the vertical direction.

From the second kinematic equation,

h2=u2*t2-(1/2)*gt2^2

where h2 is the height, t2 is the time, and u2 is the initial velocity for the second object along the vertical direction.

When the object is not in air h2=0, so using u2=uocos(θ),

0=uocos(θ)*t2-(1/2)*gt2^2

After solving the above quadratic equation, the values of t2 obtained are t2=0 which represents initial time, and t2=2uocos(θ)/g which represents the time period. So the time period of the second object is,

t2=2uocos(θ)/g

Given that the time period is the same for both cases,

t1=t2

2u1/g=2uocos(θ)/g

u1=uo cos(θ)

Calculation of the ratio of the height of the object:

The maximum height is attained when the time of the object in the air is half of the total time period. At maximum height, velocity is zero.

From the third equation of motion.

v^2=u^2-2gh

h=u^2/2g

where h is the height and u is the initial velocity of an object.

Using it and u1=uo cos(θ) and u2=uo cos(θ), the ratio of h1 and h2 is,

h1/h2= u1^2/2g÷u2^2/2g

h1/h2=(uo cos(θ))^2/(uo cos(θ))^2

h1/h2=1

The ratio of their height will be 1.

Learn more about projectile motion.

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6 0

2 years ago

The density of a sample can be obtained by dividing its __________ by its _______________.

mixer [17]

<span>Density can be calculated and found by dividing the sample's mass by its volume. D=m/v</span>

3 0

3 years ago