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3 years ago

How does the color of water affect its evaporation rate

1 answer:

8 0

Well my thinking is that the lighter the slower the water evaporates or the darker the faster becuase dark colors absorb light and allows a lot of heat to be generated or not at all.

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From the window of a house that is placed 15 m

kow [346]

Answer:

a) 52.915 m

b) The vertical velocity is approximately 21.092 m/s

The resultant velocity is approximately 26.5 m/s

Explanation:

a) The height of the window in the house from which the water was thrown = 15 m

The speed of the stream of water thrown = 20 m/s

The angle at which the water was thrown = 37° over the horizontal

The acceleration due to gravity, g = 10 m/s²

a) The distance from the base of the house at which the water will fall is given as follows;

y = y₀ + u·t·sin(θ) + 1/2·g·t²

Where;

y = The vertical height reached

u = The initial velocity

t = Time of flight

From the point the steam of water is thrown, we get;

y₀ = 15 m

Therefore;

y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²

y = 15 + 20 × t × sin(37°) - 5 × t²

When y = 0, Ground level, we get

0 = 15 + 20 × t × sin(37°) - 5 × t²

5·t² - 20×sin(37°)×t -15 = 0

∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)

t ≈ 3.3128302, or t ≈ 0.906

Therefore, the time of flight of the water, t ≈ 3.3128302 seconds

The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x

x = u·cos(θ)×t

∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915

The distance from the base of the house at which the water will fall = x ≈ 52.915 m

b) The velocity at which the water will reach the ground, 'v', is given as follows;

The vertical velocity, $v_y$ = u·sin(θ)·t - g·t

At the ground, t ≈ 3.3128302 seconds

∴ $v_y$ = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092

The vertical velocity at which the water will reach the ground, $v_y$ ≈ 21.092 m/s (downwards)

The resultant velocity, v = √( $v_y$ ² + vₓ²)

∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5

The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.

5 0

2 years ago

A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a

kumpel [21]

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

$\lambda = 720 \ nm$

or,

$= 720\times 10^{-9} \ m$

$D=0.85 \ m$

$d = 0.22 \ mm$

or,

$=0.22 \times 10^{-3} \ m$

As we know,

Fringe width is:

⇒ $\beta=\frac{\lambda D}{d}$

hence,

Separation between second and third bright fringes will be:

⇒ $\theta=\beta=\frac{\lambda D}{d}$

$=\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}$

$=2.78\times 10^{-3} \ m$

or,

$=2.78 \ mm$

8 0

2 years ago

Two groups of students were tested to compare their speed working math problems,Each group was given the same problems.One group

Zanzabum

Your hypothesis can be that the group with calculators would finish faster than the group without calculators.

4 0

3 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago

You are attempting to row across a stream in your rowboat. Your paddling speed relative to still water is 3.0 m/s (i.e., if you

Nataliya [291]

Answer:

Please check the attached file for the diagram

Explanation:

The velocity of the of the rowboat $V_{tot}$ through the river is the resultant velocity. It is obtained taking a vector sum of the velocity in still water and the velocity of the river.

There are several ways to take this vector sum, but the question makes it simple for us to use Pythagoras's theorem because the East and North directions are perpendicular to each other.

Hence;

$V_{tot}^2=V_{still}^2+V_{w}^2\\V_{tot}^2=3^2+4^2$

$V_{tot}=\sqrt{3^2+4^2}\\ V_{tot}=\sqrt{25}=5m/s$

6 0

3 years ago