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3 years ago

Which of these are part of our solar system? select all that apply

Physics

2 answers:

Len [333]3 years ago

6 0

A)planets
b)the sun
c)moons
e)comets
f)asteroids

Papessa [141]3 years ago

4 0

Everything on this list is part of the solar system EXCEPT Polaris.

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A light ray traveling through medium 1, index of refraction n1n1n_1 = 1.75, reaches the interface between medium 1 and medium 2,

horrorfan [7]

Answer:

θ₁ = 35.32°

Explanation:

given,

refractive index of medium 1 = n₁ = 1.75

refractive index of medium 2 = n₂ = 1.24

condition to describe the refracted angle

$\theta_{refracted} + \theta_{reflected}=90^0$

$\theta_2 = 90^0-\theta_1$ ...(1)

Using Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

θ₁ , θ₂ is the angle of incidence and refractive index

n₁. n₂ is the refractive index medium 1 and medium 2

1.75 x sin θ₁ = 1.24 x sin θ₂

From equation (1)

1.75 x sin θ₁ = 1.24 x sin (90-θ₁)

1.75 sin θ₁ = 1.24 cos θ₁

tan θ₁ = 0.708

θ₁ = 35.32°

Hence, angle of incidence is equal to θ₁ = 35.32°

7 0

3 years ago

Discuss potential behavioral concerns for people should they travel to Mars

Anna35 [415]

Answer:

The ability of our bodies to adapt to different levels of gravity. You would become weaker and your heart is use to zero gravity. Boredom because there isn't much to in space. When intelligent people get bored, it's not pretty all the time...

6 0

3 years ago

Who clarified the photoelectric effect?

sergiy2304 [10]

When visible light, X rays, gamma rays, or other forms of electromagnetic radiation are shined on certain kinds of matter, electrons are ejected. That phenomenon is known as the photoelectric effect. The photoelectric effect was discovered by German physicist Heinrich Hertz (1857–1894) in 1887. You can imagine the effect as follows: Suppose that a metal plate is attached by two wires to a galvanometer. (A galvanometer is an instrument for measuring the flow of electric current.) If light of the correct color is shined on the metal plate, the galvanometer may register a current. That reading indicates that electrons have been ejected from the metal plate. Those electrons then flow through the external wires and the galvanometer. HOPE THIS HELPED

7 0

3 years ago

4. A car accelerates, from rest, at 2.4 m/s?. How fast is the car traveling

pantera1 [17]

Answer:

19.2m/s

Explanation:

Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,

plug in known varibles,

v=2.4*8

v=19.2m/s

4 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago