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Alchen [17]
3 years ago
9

Which of the following is an example of a density-independent factor limiting population growth

Physics
1 answer:
Lisa [10]3 years ago
6 0
The appropriate response is the Unusual climate. These are eccentric serious or unseasonal climate; climate at the extremes of the authentic circulation—the range that has been found before. Regularly, extraordinary occasions depend on an area's recorded climate history and characterized as lying in the most strange 10%.
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5. You head downstream on a river in an outboard.
Elena-2011 [213]

Answer:

hope this helps you're welcome

5 0
3 years ago
The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many foo
sveta [45]

Answer:

It covers distance of 9.15 football fields in the said time.

Explanation:

We know that

Distance=Speed\times Time

Thus distance covered in blinking of eye =

Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters

Thus no of football fields=\frac{936}{91.4}=9.15Fields

7 0
3 years ago
How is energy sent from Earth to space?
Nostrana [21]
It might be radiation and reflection but I’m not sure
8 0
3 years ago
Read 2 more answers
Highlight two factors which shows that heat from the sun does reach the earth's surface by convection.
MAXImum [283]

radiation

when the suns radiation fall on the earth and its objects they receive heat energy and hence get heated. Thus the suns heat reaches the earth by. the process of radiation

3 0
2 years ago
A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}

now, inserting all the given values

\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

μ = 0.6

7 0
3 years ago
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