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3 years ago

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A fountain sends a stream of water straight up into the air to a maximum height of 5.60 m. The effective area of the pipe feedin

g the fountain is 4.77 x 10-4 m2. Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (1 gal = 3.79 x 10-3 m3)

1 answer:

kifflom [539]3 years ago

5 0

Answer:It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.

Read more on Brainly.com - brainly.com/question/1492542#readmore

Explanation:

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Draw a circuit that uses the resistors listed below as well as a 12V battery source. For your circuit, determine (a) the total r

Zepler [3.9K]

Answer: hello your questions lacks the required resistor values therefore i will provide a general answer using an example

answer : a) 14 ohms b) 0.86 amps c) 10.32 V

Explanation:

Assuming the resistors are : 3 ohms , 4 ohms and 5 ohms

Voltage source = 12V

<u>Assuming that the Resistors are in series </u>

<u>a) Determine Total resistance </u>

Req = R1 + R2 + R3

= 3 + 4 + 5 = 14 ohms

<u>b) Total current </u>

Ieq = V / Req

= 12 / 14 = 0.86 amps

<u>c) The Total Voltage over the entire system </u>

Vt = ∑ Voltage drops

= ( 0.86 * 3 ) + ( 0.86 * 4 ) + ( 0.86 * 5 )

= 10.32 V

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3 years ago

an electric motor converts electrical energy into which kind of energy? potential kinetic storage chemical

liubo4ka [24]

Answer:

kinetic

Explanation:

6 0

3 years ago

Read 2 more answers

A rock dropped into a pond produces a wave that takes 11.3 s to reach the opposite shore, 26.5 m away. the distance between cons

Kay [80]

The wave takes 11.3 s to cover a distance of 26.5 m, so its speed is:
$v= \frac{S}{t}= \frac{26.5 m}{11.3 s}=2.35 m/s$

The distance between two consecutive crests is 3 m, and this corresponds to the wavelength of the wave. To find its frequency, we can use the relationship between the speed v, the wavelength $\lambda$ and the frequency f:
$f= \frac{v}{\lambda}= \frac{2.35 m/s}{3 m}=0.78 Hz$

6 0

3 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

3 years ago

Three equal point charges, each with charge 2.00 μC , are placed at the vertices of an equilateral triangle whose sides are of l

oksano4ka [1.4K]

Answer:

Explanation:

Energy of system of charges

= k q₁q₂ / r₁₂ + k q₁q₃ / r₁₃ + k q₃q₂ / r₃₂

q₁ , q₂ and q₃ are charges and r₁₂ , r₁₃ , r₃₂ are densities between them

9 x 10⁹ ( 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 )

= 9 x 10⁹ x 3 x 16 x 10⁻¹²

= 432 x 10⁻³

= .432 J .

4 0

3 years ago