What is the current in a circuit with a 12.0 V battery and a 4.0 Ω resistor?

Who sided with the French

Step2247 [10]

The French were helped by the Hurons

5 0

3 years ago

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consider a charged parallel-plate capacitor. which combination of changes would quadruple its capacitance?

kogti [31]

Its capacitance Double the charge and double the plate area.

Doubling the distance between the plates of a capacitor double the capacitance. Doubling the distance between the plates of a capacitor quadruples the capacitance. As the distance between the plates decreases, the capacitance increases because the potential difference decreases.

Halving the distance between the plates of a parallel plate capacitor doubles the capacitance of the capacitor from its initial capacitance. When two or more capacitors are connected in parallel, the overall effect is that of a single equivalent capacitor with the sum of the plate areas of the individual capacitors. As we saw earlier all other factors being equal, more disk space equals more capacity.

Learn more about Capacitor here:-brainly.com/question/27393410

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4 0

2 years ago

If the angle of incidence of a light source to a shiny surface is 30 degrees, what will the angle

Virty [35]

<h3>Answer: D) 30</h3>

Angle of incidence always equals angle of reflection. Think of a tennis ball being hit into a wall. The ball will bounce off at the same angle that it approached with. The angles mentioned are formed through the line called the "normal", which is the line perpendicular to the surface.

5 0

3 years ago

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What are the names of the four hemispheres of the earth

xenn [34]

Explanation:

Northern, Southern, Eastern, and Western.

6 0

3 years ago

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

4 0

3 years ago