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3 years ago

What are some methods that astronomers use to study interstellar dust?

Physics

1 answer:

larisa [96]3 years ago

6 0

Answer:

is to observe through the telescope, study stars inside of the nebula and sometime enhance he pcture to see the real color of the nebula. hopes this helps.

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A projectile is fired at an upward angle of 35.6° from the top of a 208-m-high cliff with a speed of 170-m/s. What will be its s

sdas [7]

Answer:

Final velocity is 181.61 m/s at angle 40.44° below horizontal.

Explanation:

Initial horizontal velocity = 170 cos 35.6 = 138.23 m/s

Final horizontal velocity = 138.23 m/s

Considering vertical motion of projectile:

Initial vertical velocity, u = 170 sin 35.6 = 98.96 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = -208 m

We have v² = u² + 2as

Substituting

v² = 98.96² + 2 x -9.81 x -208

v = 117.79 m/s

Final velocity,

$v=\sqrt{138.23^2+117.79^2}=181.61m/s$

$\theta =tan^{-1}\left ( \frac{117.79}{138.23}\right )=40.44^0$

Final velocity is 181.61 m/s at angle 40.44° below horizontal.

5 0

4 years ago

Why do things become hot or cold? Relate to Thermal Energy and The Law of Conservation of Energy

Mekhanik [1.2K]

Answer:

Things become hot and cold because of the transfer of energy.

Explanation:

The energy possessed by an object or system is called thermal energy and heat is the flow of this energy. While the law of conversation of energy states that energy is not destroyed or created, it just transfers from one object to another.

When a hot object is placed in normal conditions, it transfers heat to the environment until both are at the same temperature and heat transfers from the environment to the cold objects placed in normal conditions.

4 0

3 years ago

A ray of light is moving from a material having a high indexof refraction into a material with a lower index of refraction.

luda_lava [24]

(a) Away from the normal

We can find the direction of bending of the ray of light by using Snell's equation:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where we have:

n1, n2: index of refraction of the first and second medium

$\theta_1, \theta_2$ ; angle that the incident and the refracted ray form with the normal to the surface

Here, the light ray moves from a material with high index of refraction to a material with lower index, so we have

$n_1 > n_2$

Re-arranging Snell's law we find

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

since we have

$\frac{n_1}{n_2}>1$

this implies

$sin \theta_2 > sin \theta_1\\\theta_2 > \theta_1$

so the ray of light bends away from the normal.

(b) The wavelength is greater in the second material (the one with lower index of refraction)

The wavelength of the light in a medium is given by

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda_0$ is the wavelength of the light in a vacuum

n is the refractive index

The equation can be rewritten as

$\lambda_0 = \lambda_1 n_1 = \lambda_2 n_2$

and again it can be rewritten as

$\lambda_2 = \frac{n_1}{n_2} \lambda_1$

where

$\lambda_1 = 600 nm\\\frac{n_1}{n_2}>1$

Therefore, we have that the wavelength in the second medium (the one with lower index of refraction) is longer than the wavelength in the first medium.

(c) The frequency remains the same

Wavelength and speed of a light ray depend on the medium in which the wave is travelling through, however the frequency does not depend on that, so it remains the same in the two mediums.

8 0

3 years ago

6. Calculate the pressure exerted on an elevator floor which has an

disa [49]

Answer:

Hope this helps you out if you could please leave 5 stars or brainiest thank you

3 0

3 years ago

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as

Dmitriy789 [7]

Answer:

a) $v=5.6725\,m.s^{-1}$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^{-2}$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^{-1}$

putting the values in eq. (1)

$v^2=0^2+2\times 3.62 \times 10$

$v=8.5088\,m.s^{-1}$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20\times 8.5088$

$p=170.1764\,kg.m.s^{-1}$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)\times v'=p$

$(20+10)\times v'=170.1764$

$v'=5.6725\,m.s^{-1}$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$\frac{1}{2} (M+m).v'^2=(M+m).g.h$

$\frac{1}{2} (20+10)\times 5.6725^2=(20+10)\times 9.8\times h$

$h\approx 1.6420\,m$

above the normal hanging position.

3 0

3 years ago