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hjlf
3 years ago
9

Why are the very first stars thought to have been much more massive than the sun? (a the clouds that made them were much more ma

ssive than todays star-forming clouds. (b the temperatures of the clouds that made them were higher because the clouds consisted entirely of hydrogen and helium. (c star-forming clouds were much denser early in time?
Physics
2 answers:
pashok25 [27]3 years ago
8 0
I think the correct answer would be that the temperature of the clouds that made the very first stars where thought to be higher since the clouds are made up of hydrogen and helium. Hope this answers the question. Have a nice day.
Juliette [100K]3 years ago
7 0
The main reason why the very first stars where thought to be much bigger than the sun because of the temperature of the clouds that these stars were made of were higher. The clouds that mostly form these stars consist of entirely hydrogen and helium. The answer for this would be option B.
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A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:
AfilCa [17]

The change in speed of this object is 3m/s

According to Newton's second law;

F = ma

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Given the following parameters

Force F = 8.0N

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Required

speed v

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4 0
3 years ago
An aircraft is in level flight at 225 km/hr through air at standard conditions. The lift coefficient at this speed is 0.45 and t
mojhsa [17]

Answer:

- the effective lift area for the aircraft is 8.30 m²

- the required engine thrust is 1275 N

- required power is 79.7 kW

Explanation:

Given the data in the question;

Speed V = 225 km/hr = 62.5 m/s

The lift coefficient CL = 0.45

drag coefficient CD = 0.065

mass = 900 kg

g = 9.81 m/s²

a)  the effective lift area for the aircraft

we know that for a steady level flight, weight = lift and thrust = drag

Using the equation for the lift force

F_L = C_L\frac{1}{2}ρV²A = W

we substitute

0.45 × \frac{1}{2} × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )

1081.05 × A = 8829

A = 8829 / 1081.05

A = 8.30 m²

Therefore, the effective lift area for the aircraft is 8.30 m²

b) the required engine thrust and power to maintain level flight.

we use the expression for drag force

F_D = T = C_D\frac{1}{2}ρV²A

we substitute

= 0.065 × \frac{1}{2} × 1.21 × ( 62.5 )² × 8.30

T = 1275 N

Since drag and thrust force are the same,

Therefore, the required engine thrust is 1275 N

Power required;

P = TV

p = 1275 × 62.5

p = 79687.5 W

p = ( 79687.5 / 1000 )kW

p = 79.7 kW

Therefore, required power is 79.7 kW

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