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Yuki888 [10]
3 years ago
13

Laboratory experiments here on Earth have determined that each element in the periodic table emits photons only at certain wavel

engths. These photons are apparent to astronomers as either emission or absorption lines in the spectrum of an astronomical object, like stars in galaxies. Astronomers can, by measuring the position of these spectral lines, determine which elements are present in the star itself or along the line of sight. When astronomers perform this analysis, they note that for most astronomical bodies, the observed spectral lines are all shifted to longer (redder) wavelengths. This is known as cosmological redshift and is analogous to the Doppler redshift. What does this redshift indicate to us about the universe in general?
A) The age of the universe.
B) The universe is expanding.
C) How much longer the universe should exist.
D) The universe is contracting into a black hole.
Physics
1 answer:
Bogdan [553]3 years ago
4 0

Answer: answer is B

Explanation:

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A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 358
aleksandr82 [10.1K]

Answer:

Period is 86811.5 seconds.

Explanation:

{ \boxed{ \bf{T {}^{2} =  (\frac{4 {\pi}^{2} }{GM}) {r}^{3}   }}}

{ \tt{T {}^{2}  =  \frac{4 {(3.14)}^{2} }{(6.6 \times  {10}^{ - 11} ) \times (5.98 \times  {10}^{24} )} \times  {((35880\times  {10}^{3}) } + (6370 \times  {10}^{3} )) {}^{3}   }} \\  \\ { \tt{T {}^{2}  =  7.54 \times {10}^{9} }} \\ { \tt{T =  \sqrt{7.54 \times  {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}

6 0
3 years ago
A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between
mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 × 0.00262

V = 0.4061 V

The potential difference between the plates is 0.4061 V

5 0
3 years ago
Can a body have zero velocity and finite acceleration?Explain​
sergejj [24]

Answer:

Kinda? Depends what the question is fully asking

Explanation:

Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.

So that -2 m/s thing after one second will be going -1 m/s.

After another second it'll be going 0 m/s.

After another itll be going +1 m/s and so on.

So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.

5 0
3 years ago
The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti
KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

r(t)=(8t+9)i+(2t^2-8)j+6tk

Let v is the velocity of the particle. Velocity of an object is given by :

v=\dfrac{dr(t)}{dt}

v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}

v=(8i+4tj+6k)\ m/s

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

a=\dfrac{dv(t)}{dt}

a=\dfrac{d[8i+4tj+6k]}{dt}

a=(4j)\ m/s^2

At t = 0, v=(8i+0+6k)\ m/s

v(t)=\sqrt{8^2+6^2} =10\ m/s

Hence, this is the required solution.

7 0
3 years ago
If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo
lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

temperature at t = 0

T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}

T(0) = 95°C

temperature after half hour of cooling

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

t = 30 minutes

T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}

T( 30 ) = 20 + 75 \times 0.5488

T(30) = 61.16° C

average of first half hour will be equal to

T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt

T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30

T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30

T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]

T = \dfrac{1}{30}[600 - 2058.04 + 3750]

T = 76.39°C

4 0
3 years ago
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