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3 years ago

Your low-flow showerhead is delivering water at 1.2×10−4m3/s, about 2.0 gallons per minute.

1 answer:

7 0

To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

$Q_1 = Q_2$

We know that the flow rate is equivalent to the velocity of the fluid in its area, that is,

$Q = VA$

Where

V = Velocity

A = Cross-sectional Area

Our values are given as

$Q_2 = 1.2*10^{-4}m^3/s$

$d = 0.021m$

$r = \frac{0.021}{2} = 0.0105m$

Since there is continuity we have now that,

$V_1A_1 = Q_2$

$V_1A_1 = 1.2*10^{-4}$

$V_1 = \frac{1.2*10^{-4}}{A_2}$

$V_1 = \frac{1.2*10^{-4}}{\pi r^2}$

$V_1 = \frac{1.2*10^{-4}}{\pi (0.0105)^2}$

$V_1 =0.347m/s$

Therefore the speed of the water's house supply line is 0.347m/s

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Light travels from the sun to the earth, a distance of 144,000,000,000 meters, in 480 seconds. Calculate the approximate velocit

mel-nik [20]

Answer:

The answer is C.) 300,000,000 m/s

Explanation:

Light travels at a constant rate of 300,000,000 m/s. this can be determined by dividing the distance (144,000,000,000 meters) by the time (480 seconds). that's pretty fast. in fact, nothing can travel any faster than this. ever.

5 0

3 years ago

Prove that..<br>please help<br>

GaryK [48]

$\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}$

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

$\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}$

Also,

$\large{ \longrightarrow{ \rm{ F \propto \dfrac{1}{ {d}^{2} } }}}$

Combining both, We will get

$\large{ \longrightarrow{ \rm{F \propto \dfrac{ m_1 m_2}{ {d}^{2}}}}}$

Or, We can write it as,

$\large{ \longrightarrow{ \rm{F \propto \: G \dfrac{ m_1 m_2}{ {d}^{2} }}}}$

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0

3 years ago

What is the relationship between lightning and thunder

PtichkaEL [24]

Whenever lightning strikes it separates the air where it goes. This air then rushes back together making a loud noise when it connects, creating thunder.

6 0

4 years ago

3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number

Rina8888 [55]

ANSWER

$\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}$

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

$F=\frac{GMm}{r^2}$

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

$\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}$

That is the answer.

6 0

1 year ago

It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w

Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = $\frac{mv^{2} }{r}$

Bqv = $\frac{mv^{2} }{r}$

or Eq = $\frac{mv^{2} }{r}$

Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

6 0

3 years ago