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4 years ago

Please help me out! Thanks! (:

Physics

1 answer:

ElenaW [278]4 years ago

3 0

X=Ft is a possible equation for X because by using SI units its final term became Kg metre per second

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A river with a flow rate of 7 m3/s discharges into a lake with a volume of 9,000,000 m3. The concentration of a particular VOC i

soldi70 [24.7K]

Answer:

The concentration downstream reduces to 0.03463mg/L

Explanation:

Initially let us calculate the time it is required to fill the lake, since for that period of time the pollutant shall remain in the lake before being flushed out.

Thus the detention period is calculated as

$t=\frac{V}{Q}\\\\t=\frac{9000000}{7}=1.285\times 10^{6}seconds\\\\\therefore t = 14.872days$

Now the concentration of the pollutant after 14.872 days is calculated as

$N_{t}=N_{0}e^{-kt}$

where

$N_{o}$ is the initial concentration

't' is the time elapsed after which the remaining concentration is calculated

k is the dissociation constant.

Applying values we get

$N_{t}=3\times e^{-0.3\times 14.872}$

$N_{t}=0.03463mg/L$

4 0

3 years ago

Two horizontal pipes have the same diameter, but pipe B is twice as long as pipe A. Water undergoes viscous flow in both pipes,

zheka24 [161]

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

$\Delta P = \dfrac{128 \mu L Q}{\pi D^4}$

Since the diameter in both pipes is the same, we can say:

$D = D_A = D_B$

where;

length of the first pipe A $L_A = L$ and the length of the second pipe B $L_B = 2L$

Since the difference in pressure is equivalent in both pipes:

Then:

$\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}$

$\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}$

$\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}$

$\mathbf{Q_1 = 2Q}$

6 0

3 years ago

I need C,D,E,F,H please

VARVARA [1.3K]

Answer:

cant see blury. but wanna TEXT?

Explanation:

7 0

3 years ago

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m.

Gwar [14]

Answer:

0.95 seconds

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s² (downward positive, upward negative)

Time taken by the ball to reach the maximum height

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-15}{-9.81}\\\Rightarrow t=1.52\ s$

Maximum height

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-15^2}{2\times -9.81}\\\Rightarrow s=11.47\ m$

Distance between maximum height of the ball and the branch is 11.47-7 = 4.46 m

So, the distance that will be covered on the way down is 4.46 m

Now

u = 0

s = 4.46

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.46=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4.46\times 2}{9.81}}\\\Rightarrow t=0.95\ s$

Time taken by the ball from the maximum height to the tree branch is 0.95 seconds.

Total time taken from the moment the ball is thrown to reach the tree branch is 1.52+0.95 = 2.47 seconds

7 0

3 years ago

The mean distance from Saturn to the sun is 9 times greater than the mean distance from Earth to the sun. How long is a Saturn y

Diano4ka-milaya [45]

It probably also has a nineth of the speed, so it's 81 earth years. If they had the same speed, it would be 9 earth years. Probably the first one is more accurate :-??

3 0

3 years ago