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3 years ago

What Species can change over time to adapt to their environment

Physics

1 answer:

ryzh [129]3 years ago

3 0

Answer:

Explanation:

Most environments have many niches. If a niche is "empty" (no organisms are occupying it), new species are likely to evolve to occupy it. This happens by the process of natural selection. By natural selection, the nature of the species gradually changes to become adapted to the niche.

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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

AVprozaik [17]

Answer:

v=115 m/s

v=414 km/h

Explanation:

Given data

$A_{area}=0.140m^{2}\\ p_{air}=1.21 kg/m^{3}\\ m_{mass}=80kg$

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

$mg=1/2*C*p_{air}*v^{2}*A_{area}\\ v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s$

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

$v=(115*3.6)km/h\\v=414km/h$

5 0

3 years ago

Which of the following is the best reason that trees care important for overall air quality?

Marina86 [1]

Trees are important because oxygen

4 0

3 years ago

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.

From trigonometry, $\sin A =\cos B$ if A and B are complementary.

At $t = 0$ , $x = 3.5$

$3.5 = A\cos(\omega \times0)$

$A =3.5$

$x = 3.5\cos(\omega t)$

At $t = 0.12$ , $x = 1.5$

$1.5 = 3.5\cos(0.12\omega)$

$\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286$

$0.12\omega =\cos^{-1}0.4286$

$0.12\omega = 1.13$

$\omega = 9.4$

The period, $T$ , is related to $\omega$ by

$T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67$

5 0

3 years ago

A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and

Troyanec [42]

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ F = N - W = 0

• net horizontal force:

∑ F = Fs = m a

where

N = magnitude of normal force

W = car's weight

Fs = mag. of static friction

m = car's mass

a = v ²/R = mag. of the centripetal acceleration

v = car's speed

R = radius of curve

Now,

• compute the car's weight:

W = m g = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

N = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

Fs = 0.5 N = 7350 N

• solve for the (maximum) acceleration:

7350 N = (1500 kg) a → a = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = v ²/ (35 m) → v ≈ 13 m/s

4 0

3 years ago

A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 5.9 m/s. if the coefficient of fricti

bagirrra123 [75]

Frictional force = vertical force x coeff of friction = 590N x 0.045 = 26.55N

Mass of boy and sled = 590N / g = 590N / 9.8m/s^2 = 60.20 kg

Deceleration due to friction = 26.55N / 60.20kg = 0.44 m/s^2.

For constant acceleration we have:

v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and ti is time. In this case v = 0, u = 5.9 m/s, a = -0.44 m/s^2. So we have
0 = 5.9 - 0.44t
which gives t = 5.9 / 0.44 = 13.409 s.

Distance traveled in this time d = ut + 0.5at^2 = 5.9 x 13.409 - 0.5 x 0.44 x 13.409^2 = 39.56 m

Answer: 40 meters

8 0

3 years ago