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4 years ago

Describe how the amount of voltage applied to a circuit affects current flow.

Physics

2 answers:

posledela4 years ago

8 0

If you’re on Usa Test Prepnor something, the answer will be C

pychu [463]4 years ago

5 0

Increasing the voltage will increase the current proportionately.

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2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind

Alex777 [14]

The resultant speed of the plane is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

$ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}$

$ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}$

Hence, we have that the resultant speed of the plane is (3) 226 m/s

Have a nice day!

5 0

3 years ago

A skater has rotational inertia 4.2 kg-m2 with his fists held to his chest and 5.7 kg?m2 with his arms outstretched. The skater

Valentin [98]

Answer: 38.5rad/s

Explanation: The calculations can be viewed on the image attached below. Thanks

4 0

3 years ago

While taking a shower, you notice that the shower head is made up of 44 small round openings, each with a radius of 2.00 mm. You

Julli [10]

To solve the problem it is necessary to apply the concepts related to the calculation of discharge flow, Bernoulli equations and energy conservation in incompressible fluids.

PART A) For the calculation of the velocity we define the area and the flow, thus

$A = \pi r^2$

$A = pi (2*10^{-3})^2$

$A = 12.56*10^{-6}m^2$

At the same time the rate of flow would be

$Q = \frac{1L}{2s}$

$Q = 0.5L/s = 0.5*10^{-3}m^3/s$

By definition the discharge is expressed as

$Q = NAv$

Where,

A= Area

v = velocity

N = Number of exits

Q = NAv

Re-arrange to find v,

$v = \frac{Q}{NA}$

$v = \frac{0.5*10^{-3}}{44*12.56*10^{-6}}$

$v = 0.9047m/s$

PART B) From the continuity equations formulated by Bernoulli we can calculate the speed of water in the pipe

$P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 +\frac{1}{2}\rho v^2_2 +\rho g h_2$

Replacing with our values we have that

$1.5*10^5 + \frac{1}{2}(1000) v_1^2+(1000)(9.8)(0) = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)$

$v_1^2 = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)$

$v_1 = \sqrt{10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)}$

$v_1 = 3.54097m/s$

PART C) Assuming that water is an incomprehensible fluid we have to,

$Q_{pipe} = Q_{shower}$

$v_{pipe}A_{pipe}=v_{shower}A_{shower}$

$3.54097*A_{pipe}=0.9047*12.56*10^{-6}$

$A_{pipe} = \frac{0.9047*12.56*10^{-6}}{3.54097}$

$A_{pipe = 3.209*10^{-6}m^2$

3 0

3 years ago

Bina satu pernyataan benar dan satu pernyataan palsu bagi setiap

Papessa [141]

Answer:

which language is this?

I do not understand

Explanation:

5 0

4 years ago

Magnetic field strength decreases as you get farther from the poles of the magnet. does the flux through the face of the coil ch

brilliants [131]

Magnetic field strength decreases as you get farther from the poles of the magnet.

<h3>Explanation</h3>

There are three ways to change the magnetic flux across a loop: Change the strength of the magnetic field across the surface (raise, reduce).

It doesn’t matter whether you move the coil while keeping the magnet fixed or the other way around if you want to change the surface area of the loop (raise by expanding the loop, decrease by reducing the loop).

The way they move in respect to one another determines how the magnetic flux across the coil changes. The direction in which the galvanometer needle deflects depends on which side of the coil confronts the magnet when you move it. The coil ends are labelled to show which goes to the red wire and which goes to the black wire.

To know more about magnetic flux visit:

brainly.com/question/16960581

#SPJ4

5 0

2 years ago