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allochka39001 [22]
3 years ago
15

What is a volt as applied in physics ​

Physics
1 answer:
Alexus [3.1K]3 years ago
8 0

Answer:

The answer is below

Explanation:

Volt in physics is a term that is used to describe the unit of both electrical potential difference and electromotive force. In other words, it is a unit of measuring the voltage between two points. It is represented as "V".

The formula for Voltage is V=IR

Where V= Voltage, I = Current and R = Resistance.

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A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with
Julli [10]

Explanation:

Given that,

Capacitor C=1.66\ \mu F

Resistor R=80.0\ \Omega

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

f_{c}=\dfrac{1}{2\pi R C}

Where, R = resistor

C = capacitor

Put the value into the formula

f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}

f_{c}=1198.45\ Hz

(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})

Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=2.280\ Volt

(C). We need to calculate the V_{R} when f = f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=3.606\ Volt

(D). We need to calculate the V_{R} when f = 2f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

Hence, This is the required solution.

8 0
3 years ago
Find the magnitude of this<br> vector:<br> 174 m<br> N<br> 188.4 m<br> HELP FAST
Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

a^2+b^2=c^2

Plug values in

88.4^2+174^2=c^2

Simplify

7814.56+30276=c^2

Add the two values

38090.56=c^2

Take the square root of both sides

195.168\approx195.168

8 0
2 years ago
The type of lens that spreads out parallel light is a
ohaa [14]

the answer is concave lens

7 0
3 years ago
How does an environmental change (ex. hurricane, drought, etc.) affect the organisms that are adapted to the environment before
rjkz [21]
The environmental changes can will negatively affect different organisms differently, for example some organisms might die due to failure of the organisms to adapt to the new changes. Some organisms will be forced to migrate in order to counter these changes, this may lead to this organisms leaving their old living routine and adapt to the new methods of living, this may lead to evolution caused by new environment.  
3 0
3 years ago
Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass thro
blagie [28]

Answer:

21.2\times 10^{-6} T

Explanation:

i  = magnitude of current in each wire = 2.0 A

a  = length of the side of the square = 4 cm = 0.04 m

r  = length of the diagonal of the square = \sqrt{2} a = \sqrt{2} (0.04) = 0.057 m

B = magnitude of magnetic field by wires at A and C

B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )

B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )

B = 10\times 10^{-6} T

B' = magnitude of magnetic field by wire at B

B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )

B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )

B' = 7.02\times 10^{-6} T

Net magnitude of the magnetic field at D is given as

B_{net} = \sqrt{B^{2}+B^{2}} + B'

B_{net} = \sqrt{2} B + B'

B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})

B_{net} = 21.2\times 10^{-6} T

8 0
3 years ago
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