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3 years ago

Photographs of many young stars show long jets of material apparently being ejected from their poles.

Physics

1 answer:

Rudiy273 years ago

7 0

Answer:

A) True

Explanation:

Researchers have detected numerous jets of gas ejected from poles of young stars and planetary nebulae.

By examining images of hydrogen molecules excited at infrared wavelengths, scientists have been able to see through the gas and dust in the Milky Way, in order to observe the most distant targets. These goals are normally hidden from view and many of them have never been seen before.

The entire study area covers approximately 1,450 times the size of the full moon, or the equivalent of an image of 95 gigapixels. The survey reveals jets emanating from proto-stars and planetary nebulas, as well as remnants of supernovae, the illuminated edges of vast clouds of gas and dust, and the warm regions that surround massive stars and their associated groups of smaller stars.

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Assume a satellite shines an unpolarized light on a telescope. The intensity of the light as it reaches the telescope is 1.1*10-

n200080 [17]

Answer:

$4.125\times 10^{-11}\ W/m^2$

Explanation:

$I_0$ = Intensity of unpolarized light = $1.1\times 10^{-10}\ W/m^2$

$\theta$ = Angle of the filter = $30^{\circ}$

Intensity of light is given by

$I=\dfrac{I_0}{2}cos^2\theta\\\Rightarrow I=\dfrac{1.1\times 10^{-10}}{2}cos^230\\\Rightarrow I=4.125\times 10^{-11}\ W/m^2$

The intensity of light detected by the camera is $4.125\times 10^{-11}\ W/m^2$

7 0

3 years ago

In a 49 s interval, 595 hailstones strike a glass window of an area of 0.954 m at an angle of 25° to the window surface. Each ha

eduard

Average force on the window: 0.32 N

Explanation:

The average force exerted on the window is given by Newton's second law

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the net change in momentum of the hailstones in a time interval of $\Delta t$

In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:

$p_i =m u sin \theta$ is the initial momentum of one hailstone, with

m = 7 g = 0.007 kg is the mass

$u=4.5 m/s$ is the initial speed

$\theta=25^{\circ}$ is the angle with the window

The final momentum is

$p_f = mv sin \theta$

where

v = 4.5 m/s is the final speed (the collision is elastic so the speed is equal, while the direction changes)

$\theta=-25^{\circ}$ (after the rebound, the direction has changed)

So the change in momentum of 1 hailstone is

$\Delta p = mv sin(-25^{\circ})-mu sin(25^{\circ})=-2mu sin(25^{\circ})=-0.0266 kg m/s$

We are interested only in the magnitude, so

$\Delta p = 0.0266 kg m/s$

There are 595 hailstones hitting the window in 49 s, so the total change in momentum is

$\Delta p = 595\cdot 0.0266 = 15.8 kg m/s$

And therefore, the average force on the window is

$F=\frac{\Delta p}{\Delta t}=\frac{15.8}{49}=0.32 N$

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3 0

2 years ago

An fm radio station broadcasts at 98. 6 mhz. What is the wavelength of the radiowaves?.

victus00 [196]

The wavelength of the radio waves is 3.04 cm.

<h3>Calculation:</h3>

λf = c

λ = c/f

where,

λ = wavelength

c = speed of light

f = frequency

Given,

f = 98.6 MHz = 98.6 × 10⁶

c = 3 × 10⁸

To find,

λ =?

Put the values in the formula,

λ = c/f

λ = 3 × 10⁸/98.6 × 10⁶

= 0.0304 × 10² m

= 3.04 cm

Therefore, the wavelength of the radio waves is 3.04 cm.

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3 0

1 year ago

At what point on the hill will the car have zero gravitational potential energy?

Mila [183]

At the bottom of the hill

8 0

2 years ago

Read 2 more answers

suppose a vacuum cleaner use 120 j of electrical energy. If 45 j are used to pull air into the vacuum cleaner, how efficient is

Mice21 [21]

It depends on how much time 45 j will go, so if you told me that it went through 45 j per minute it will go through for 2 minutes so not efficient unless it went through 45 j per hour then it is efficient.

8 0

3 years ago