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barxatty [35]
3 years ago
9

"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in gro

ups dozens of millions of lightyears away. What method do astronomers use with the Hubble to find such distances
Physics
1 answer:
pickupchik [31]3 years ago
5 0

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called  finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

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Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st
scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

x(t) = v_0 cos \theta t

where

v_0 is the initial speed

t is the time

\theta=32.0^{\circ} is the angle below the horizontal

We can rewrite this equation as

t=\frac{x(t)}{v_0 cos \theta} (1)

The vertical displacement instead is given by

y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2 (2)

where

g=9.8 m/s^2 is the acceleration of gravity

Substituting (1) into (2),

y(t) = -x(t) tan \theta - \frac{1}{2}gt^2

We know that for t = time of flight, the horizontal displacement is

x(t) =50.8 m

We also know that the vertical displacement is

y(t) = -45 m

Substituting everything into the equation, we can find the time of flight:

\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

x(t) = v_0 cos \theta t

where we have

x = 50.8 m

\theta=32.0^{\circ}

Substituting the time of flight,

t = 1.64 s

We find:

v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s

The initial vertical velocity is instead

u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s

And it changes according to the equation

v_y = u_y -gt

So at t = 1.64 s (when the ball hits the ground),

v_y = -19.3 - (9.8)(1.64)=-35.4 m/s

So the impact speed is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s

While the direction is:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}

8 0
3 years ago
Which situations will NOT increase the reactions rate of most chemical reactions?
kap26 [50]
I think B I’m guessing
5 0
3 years ago
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A girl with a mass of 27 kg is playing on a swing. There are three main forces
N76 [4]

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

  • Mass of the girl, m = 27 kg
  • Speed of the girl, v = 3 m/s
  • Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

3 0
2 years ago
A ford F-250 pulls a 1000-kg car with a net force of 2000 N. What is the acceleration of the car A=F/m
vfiekz [6]

Answer:

2ms^2

Explanation:

A = 2000 divided by 1000

A= 2

5 0
3 years ago
Why sound heard better at night than at day?​
vodomira [7]

Answer:

it is because it is quieter at night than in the daytime. Therefore it is easy to hear the sound far away. However, it is only one of the reasons. Actually, sound transmits farther at night may be related to refraction of sound waves! First, sound is the vibration of air, and it is a kind of wave motion.

Explanation:

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3 years ago
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