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o-na [289]
3 years ago
11

Which of the following is the main evidence of life in the early universe?

Physics
2 answers:
Arturiano [62]3 years ago
8 0
A) Dinosaur fossils because they were proven that there was life before us. we also found out that our old skulls had dinosaur teeth in the heads. that showed us that we were on the food chain. but now we aren't because we now have protected cities.
Assoli18 [71]3 years ago
3 0

Answer:

Cyanobacteria.

Explanation:

The oldest forms of life would be considered the cyanobacteria from the options, this bacteria are called extremists because they can survive in environments with littlo to zero oxygen andis thought that they have been present from the early begining of the universe because they are the only living organisms known to mankind that are able to survive under those conditions.

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What is the mass of an object moving with 80N of force and an acceleration of 8 m/s2?
OverLord2011 [107]

Answer:

10 Kg

Explanation:

Force is equal to mass times acceleration

therefore mass is equal to force divided by acceleration

please mark me brainliest

6 0
2 years ago
Calculate the work done by the cyclist when his power output is 200 W for 1800 seconds.
Kobotan [32]

Answer:

3.6 × 10^5 J

Explanation:

Work Done = Energy

Energy = Power × time

Energy = 200 × 1800

= 360000 J

W =

3.6 \times 10 {}^{5} j

3 0
3 years ago
An incompressible fluid flows at .252 m/s through a 44 diameter (circular cross section) pipe. The pipe widens to a square cross
Serggg [28]

ANSWER:

The easiest way to get a fairly accurate measure of your water flow rate is to time yourself filling up a bucket. So for example if you fill up a 10 litre bucket in 1.5 minutes, then your flow rate will be: 10/1.5 = 6.66 Litres per minute.

7 0
2 years ago
What must be the acceleration of a train in order for it to stop 12 m/s in a distance if 541 m
earnstyle [38]

Answer:

The acceleration of the train must be - 0.133 m/s²

Explanation:

Lets explain how to solve the problem

A train in order for it to stop 12 m/s in a distance if 541 m

That means the initial velocity of the train is 12 m/s

Its final velocity is zero (stop)

The distance it covers is 541 m

We want to find its acceleration

The acceleration will be negative quantity because the train reduced its

velocity from 12 m/s to zero

We need rule contains velocity, acceleration and distance

So we will use ⇒<em> v² = u² + 2as</em>, where v is the final velocity, u is the

initial velocity, a is the acceleration and s is the distance

v = 0, u = 12 m/s, s = 541 m

Substitute these values in the rule

(0)² = (12)² + 2(a)(541)

0 = 144 + 1082 a

Subtract 144 from both sides

-144 = 1082 a

Divide both sides by 1082

- 0.133 = a

<em>The acceleration of the train must be - 0.133 m/s²</em>

3 0
3 years ago
NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0
3 years ago
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