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Molodets [167]
1 year ago
8

The main difficulty with multiple star systems such as Alpha Centauri as places to look for life is that: _________

Physics
1 answer:
Leona [35]1 year ago
8 0

Answer:

The conditions for life are extremely specific and the search is extremely broad.

Explanation:

in order to sustain life, the planet has to have water.

Too hot and the water will evaporate too cold and the water will freeze, this is based on the planet's distance from the sun and its solar mass. (The mass compared to our sun) If the solar mass is double that of our sun then the planet would have to be double the distance from the sun in order to sustain life and keep the water from evaporating. Let me Know if This helps

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Suppose the earth is shaped as a sphere with radius 4,0004,000 miles and suppose it rotates once every 24 hours. How many miles
Alenkinab [10]

Answer:

1047 miles

Explanation:

The radius of the Earth is

r = 4000 (miles)

So its circumference, which is the total length of the equator, is given by

L=2\pi r= 2\pi(4000)=25133 mi

Now we know that the Earth rotates once every 24 hours. So the distance through which the equator moves in one hour is equal to its total length divided by the number of hours, 24:

L' = \frac{25133 mi}{24h}=1047 mi

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3 years ago
How can you turn off a magnetic field produced by an electrical current?
aleksklad [387]

Answer:By turning the electrical current off

Explanation:Trust me I took the test

6 0
3 years ago
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A bowling ball is pushed with a force of 22.0N and accerlates at 5.5 m/s square. What is the mass of the bowling ball
stealth61 [152]
Mass of bowling ball is 4.0kg (A)
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2 years ago
Which of the following is an example of a force?
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7 0
2 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

8 0
3 years ago
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