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1 year ago

5

The water hose at your home outputs 24 gallons of water a minute. how many cubic centimeters per second does the hose output? th

ere are 264.172 gal in 1 m3 and 100 cm in 1 m.

1 answer:

adoni [48]1 year ago

7 0

When we convert 24 gallon/min to cubic centimeters per second (cm³/s), the result obtained is 1514.164 cm³/s

<h3>How to convert 24 gallon/min to cm³/min</h3>

We'll begin by converting 24 gallon/min to cm³/min. This can be obtained as follow:

1 gallon/min = 3785.41 cm³/min

Therefore,

24 gallon/min = (24 gallon/min × 3785.41 cm³/min) / 1 gallon/min

24 gallon/min = 90849.84 cm³/min

<h3>How to convert 90849.84 cm³/min to cm³/s</h3>

We can convert 90849.84 cm³/min to cm³/s as follow:

60 cm³/min = 1 cm³/s

Therefore,

90849.84 cm³/min = (90849.84 cm³/min × 1 cm³/s) / 60 cm³/min

90849.84 cm³/min = 1514.164 cm³/s

Thus,

24 gallon/min = 1514.164 cm³/s

Learn more about conversion:

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The propeller of an airplane is at rest when the pilot starts the engine; and its angular acceleration is a constant value. Two

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Answer:0.318 revolutions

Explanation:

Given

Initially Propeller is at rest i.e. $\omega _0=0 rad/s$

after $t=10 s$

$\omega =10 rad/s$

using $\omega =\omega _0+\alpha t$

$10=0+\alpha \cdot 10$

$\alpha =1 rad/s^2$

Revolutions turned in 2 s

$\theta =\omega _0t+\frac{\alpha t^2}{2}$

$\theta =0+\frac{1\times 2^2}{2}$

$\theta =2 rad$

To get revolution $\frac{\theta }{2\pi }$

= $\frac{2}{2\pi}=0.318\ revolutions$

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3 years ago

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By what factor is the self-inductance of an air solenoid changed if its length and number of coil turns are both tripled

fredd [130]

Answer:

The new self inductance is 3 times of the initial self inductance.

Explanation:

The self inductance of a solenoid is given by :

$L=\dfrac{\mu_oN^2 A}{L}$

Where

N is number of turns per unit length

A is area of cross section

l is length of solenoid

If length and number of coil turns are both tripled,

l' = 3l and N' = 3N

New self inductance is given by :

$L'=\dfrac{\mu_oN'^2 A}{L'}\\\\=\dfrac{\mu_o(3N)^2 A}{3L}\\\\=3\dfrac{\mu_oN^2 A}{L}\\\\=3L$

So, the new self inductance is 3 times of the initial self inductance.

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2 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

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3 years ago

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Answer:

Fy=107.2 N

Explanation:

Conceptual analysis

For a right triangle :

sinβ = y/h formula (1)

cosβ = x/h formula (2)

x: side adjacent to the β angle

y: opposite side of the β angle

h: hypotenuse

Known data

h = T = 153.8 N : rope tension

β= 44.2°with the horizontal (x)

Problem development

We apply the formula (1) to calculate Ty : vertical component of the rope force.

sin44.2° = Ty/153.8 N

Ty = (153.8 N ) *(sen44.2°)= 107.2 N directed down

for equilibrium system

Fy= Ty=107.2 N

Fy=107.2 N upward component of the force acting on the stake

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2 years ago