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3 years ago

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A population of birds walks along the shores of a swampy area eating small crabs. Climatic changes cause the water level in the

area to increase. Which genetic variation in the bird population will be at an advantance leading to greater survival in the area?

1 answer:

sergejj [24]3 years ago

4 0

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Use the worked example above to help you solve this problem. An airboat with mass 4.30 102 kg, including the passenger, has an e

umka21 [38]

Answer:

a) a = 1,865 m / s² and b) t = 8.1 s

Explanation:

a) Let's use Newton's second law to find acceleration, we can work the equation in scalar form because displacement and force have the same direction

F = m .a

a = F / m

a = 8.02 10² /4.3 10²

a = 1,865 m / s²

b) We use kinematic relationships in one dimension

vf = vo + at

vf = 0 + a t

t = vf / a

t = 15.1 / 1.865

t = 8.1 s

3 0

3 years ago

What factors affect the gravitational pull of two massive objects?

Mrrafil [7]

Answer:Mass and distance

Explanation:The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.

Hope this helps!!

3 0

3 years ago

Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside

jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

$\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}$

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

$\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}$

Thus, the flow rate of the water at the later position is 5.99 m/s

4 0

1 year ago

Two astronauts of mass 100 kg are 2 m apart in outer space. What is the

fredd [130]

The force of gravity between the astronauts is $1.67\cdot 10^{-7}N$

Explanation:

The magnitude of the gravitational force between two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have two astronauts, whose masses are:

$m_1 = 100 kg\\m_2 = 100 kg$

While the separation between the astronauts is

r = 2 m

Substituting into the equation, we can find the gravitational force between the two astronauts:

$F=\frac{(6.67\cdot 10^{-11})(100)(100)}{2^2}=1.67\cdot 10^{-7}N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

Adam uses a fixed pulley, such as the one shown below, to lift an object. Adam applies an input force to the pulley as he pulls

stira [4]

When Adam applies a ‘pull’ force on the pulley, there is an output force that the pulley lets out, directly pulling the object with it. We cannot always pull up objects with our bear hands, no matter how much force we apply. Which is why pulleys allow us to apply the force and pulleys do the work of pulling the objects for us, since work and force come hand in hand.

6 0

3 years ago