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3 years ago

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Suppose that the habitat of a species that once lived on land has now become covered in water. In order to survive, the species

adapts to have webbed feet. This adaptation likely takes place A.within a week of the habitat change. B.after a single generation. C.immediately. D.over millions of years.

2 answers:

netineya [11]3 years ago

3 0

D.over millions of years

Nutka1998 [239]3 years ago

3 0

Answer:

D.over millions of years

Explanation:

That was the answer in study island :/

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A 81 kg man is riding on a 40 kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed relative to the g

Alexus [3.1K]

Answer:

$\Delta v= 4.66\frac{m}{s}$

Explanation:

In this case we have to use the Principle of conservation of Momentum:

<em>This principle says that in a system the total momentum is constant if no external forces act in the system. The formula is:</em>

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

<em>Where:</em>

$m_1:$ Mass of the first object.

$m_2:$ Mass of the second object.

$v_1:$ Initial velocity of the first object.

$v_2:$ Initial velocity of the second object.

$u_1:$ Final velocity of the first object.

$u_2:$ Final velocity of the second object.

In <u>this problem</u> we have:

$m_1=81kg\\m_2=40kg\\v_1_2=2.3\frac{m}{s}$

$u_1=0\frac{m}{s}$

Observation: $v_1_2:$ Is because the system has the same initial velocity.

First we have to find $u_2$ ,

$m_1v_1+m_2v_2=m_1u_1+m_2u_2$

We can rewrite it as:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2$

Replacing with the data:

$(m_1+m_2)v_1_2=m_1u_1+m_2u_2\\\\(81kg+40kg)2.3\frac{m}{s}=81kg(0\frac{m}{s})+40kg(u_2)\\\\(121kg)2.3\frac{m}{s}=40kg(u_2)\\\\\frac{(121kg)2.3\frac{m}{s}}{40kg}=u_2\\\\\frac{278.3}{40}\frac{m}{s}=u_2\\\\6.96\frac{m}{s}=u_2$

We found the final velocity of the cart, but the problem asks for the resulting change in the cart speed, this means:

$\Delta v=u_2-v_2\\\Delta v=6.96\frac{m}{s}-2.3\frac{m}{s}\\\Delta v= 4.66\frac{m}{s}$

Then, the resulting change in the cart speed is:

$\Delta v= 4.66\frac{m}{s}$

5 0

3 years ago

Sound waves travel faster through air than through solids.<br> a. True<br> b. False

WINSTONCH [101]

I believe the answer is true

7 0

3 years ago

PLEASE HELP AS SOON AS POSSIBLE!!!

Brums [2.3K]

Answer:

h = 81.63 m

Explanation:

Given that,

The speed of the car, v = 40 m/s

We need to find the height when the car comes to rest. We can use the conservation of energy to find it i.e.

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}\\\\h=\dfrac{(40)^2}{2\times 9.8}\\\\h=81.63\ m$

So, it will reach to a height of 81.63 m and comes to rest.

7 0

3 years ago

A 5 m3 tank containing 5kg of an unknown ideal gas at 500 kPa is connected through a valve to another tank containing 10 kg of t

Ivan

Answer:

a) $V_{T} = 9\,m^{2}$ , b) $m_{T} = 15\,kg$ , c) $P_{T} = 416.667\,kPa$

Explanation:

a) The equation of state for ideal gas is:

$P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

Given the existence of an isothermal process, the following relation is derived:

$\frac{P_{1}\cdot V_{1}}{m_{1}} = \frac{P_{2}\cdot V_{2}}{m_{2}}$

The volume of the other tank is:

$V_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}$

$V_{2} = \left(\frac{10\,kg}{5\,kg} \right)\cdot \left(\frac{200\,kPa}{500\,kPa}\right)\cdot (5\,m^{3})$

$V_{2} = 4\,m^{3}$

The total volume is:

$V_{T} = V_{1} + V_{2}$

$V_{T} = 5\,m^{3} + 4\,m^{3}$

$V_{T} = 9\,m^{2}$

b) The total mass is:

$m_{T} = m_{1} + m_{2}$

$m_{T} = 5\,kg + 10\,kg$

$m_{T} = 15\,kg$

c) The pressure of the gas in the two tanks is:

$P_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{V_{1}}{V_{2}}\right)\cdot P_{1}$

$P_{T} = \left(\frac{15\,kg}{5\,kg}\right)\cdot \left(\frac{5\,m^{2}}{9\,m^{2}} \right)\cdot (500\,kPa)$

$P_{T} = 416.667\,kPa$

3 0

4 years ago

Scorpion4ik [409]

Answer:

SKID

Explanation:

In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.

Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.

Y axis y

N- W = 0

N = W

X axis (radial)

fr = m a

the acceleration in the curve is centripetal

a = $\frac{v^2}{r}$

the friction force has the expression

fr = μ N

we substitute

μ mg = m v²/r

v = $\sqrt{\mu g r}$

we calculate

v = $\sqrt{0.1 \ 9.8 \ 3}$

v = 1,715 m / s

to compare with the cyclist's speed let's reduce to the SI system

v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s

We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID

5 0

3 years ago