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Novay_Z [31]
2 years ago
7

The universe was 5 percent its current size when light left objects observed now at redshift of ______________

Physics
1 answer:
Nuetrik [128]2 years ago
4 0

The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.

  • The Universe was 5 percent its current size when light left objects now at redshift of <u>19</u>.

Reasons:

The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

\displaystyle \frac{a}{a_0}  =\mathbf{ \frac{1}{1 + z}}

Where;

a₀ = The current size of the Universe

a = The size of the early Universe = 5% of a

Therefore;

\displaystyle \frac{a}{a_0}  =5\% = 0.05=  \frac{1}{1 + z}

\displaystyle 0.05  = \frac{1}{1 + z}

0.05 + 0.05·z = 1

\displaystyle z = \mathbf{ \frac{1 - 0.05}{0.05} } = 19

  • The redshift is of the observed light is, z = <u>19</u>

Learn more here:

brainly.com/question/14459434

brainly.com/question/3654558

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The actual mechanical advantage of any machine is its__________divided by its________________
zaharov [31]

Answer: Load divided by it effort

Explanation:

Mechanical advantage of any machine is its load divided by its effort

4 0
3 years ago
When you drive through deep water, you should dry out your brakes by _____________. pumping your brakes and driving in low gear
Mandarinka [93]

Answer:

By Applying pressure to the brakes

Explanation:

Driving cars through deep water that is more than 10cm can make the cars to float. Most modern cars are usually water- tight so they can start to float through water that is about 30cm deep, fast moving water is very powerful so one needs to be very careful when driving.

If the brakes are wet test them by pressing or tapping on them gently.

You can as well dry brakes by driving in low gear and applying pressure to the brakes.

7 0
2 years ago
I need help for the first 3 plz.
madam [21]
1) hypothesis
2) data
3) method

I think these are correct.
7 0
2 years ago
Read 2 more answers
A 16.0kg canoe moving to the left at 12.5m/s makes an elastic head-on collision with a 14.0kg raft moving to the right at 16.0m/
kherson [118]

The canoe is moving at 14.1 m/s to the right after the collision.

Explanation:

According to the law of conservation of momentum, in absence of external forces the total momentum of the system must be conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 16.0 kg is the mass of the canoe

u_1 = -12.5 m/s is the initial velocity of canoe (we take right as positive direction, and since the canoe is moving to the left, its velocity is negative)

v_1 is the final velocity of the canoe

m_2 = 14.0 kg is the mass of the raft

u_2 = +16.0 m/s is the initial velocity of the raft

v_2 = -14.4 m/s is the final velocity of the raft

Re-arranging the equation and substituting the values, we find: the final velocity of the canoe:

v_1 = \frac{m_1 u_1 + m_2 u_2-m_2 v_2}{m_1}=\frac{(16.0)(-12.5)+(14.0)(16.0)-(14.0)(-14.4)}{16.0}=+14.1 m/s

So, the canoe is moving at 14.1 m/s to the right after the collision.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

5 0
3 years ago
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0
2 years ago
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