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4 years ago

A power source of 2.0 V is attached to the ends of a capacitor. The capacitance is 4.0 μF. What is the amount of charge stored i

n this capacitor?

Physics

1 answer:

ioda4 years ago

8 0

$C = 4 \ \mu F = 4 \cdot 10^{-6} \ F. \newline
q = Cu = 4 \cdot 10^{-6} \cdot 2 = 8 \cdot 10^{-6} = 0.000008 \ C.$

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A motorist drives south at 24.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.20 min, and finally travels nort

grigory [225]

Answer:

magnitude m = 5440.26 M

average speed is = 25.05 m/s

average velocity = 14.62 m/s

Explanation:

given data

speed v1 = 24 m/s

time t1 = 3 min = 180 s

speed v2 = 25 m/s

time t2 = 2.2 min = 132 s

speed v3 = 30 m/s

time t3 = 1 min = 60 s

total time = 6.2 min = 372 s

to find out

displacement in m and average speed and average velocity

solution

so we find here displacement

displacement = vt .............1

displacement 1 = 24 × 180 = 4220 m south = -4220 j ..........2

displacement 2 = 25 × 132 = 3300 m west = -3300 i ...........3

distance 3 = 30 × 60 = 1800 m

displacement = 1800 cos45 -i + 1800 cos45 j

displacement = -1272.79 i + 1272.79 j .............4

so total displacement D = equation 2 + equation 3 + equation 4

total displacement D = -4220 j -3300 i -1272.79 i + 1272.79 j

total displacement D = -2947.21 j - 4572.79 i

so magnitude m = √(-2947.21)² + (-4572.79)² )

so magnitude m = 5440.26 M

and

average speed is = total distance / time

average speed is = (4220 + 3300+ 1800 ) / 372

average speed is = 25.05 m/s

and

average velocity = total displacement / time

average velocity = 5440.26 / 372

average velocity = 14.62 m/s

7 0

4 years ago

A guitar string is supposed to have a fundamental frequency 256 Hz. It currently has a fundamental frequency 248 Hz when the str

umka2103 [35]

Answer:

The tension to bring the guitar string into tune is 372.95 Hz.

Explanation:

Given;

current frequency, f₁ = 248 Hz

current tension, T₁ = 350 N

fundamental frequency, f₂ = 256

The tension on the string to bring the guitar string into tune is calculated as;

$v = \sqrt{\frac{T}{\mu} } \\\\f\lambda = \sqrt{\frac{T}{\mu} } \\\\f^2\lambda^2 = \frac{T}{\mu} \\\\f^2 = \frac{T}{\mu \lambda^2}\\\\let \ {\mu \lambda^2} = k\\\\f^2 =\frac{T}{k} \\\\k = \frac{T}{f^2} \\\\\frac{T_1}{f_1^2} = \frac{T_2}{f_2^2}\\\\T_2 = \frac{T_1 f_2^2}{f_1^2} \\\\T_2 = \frac{350 \times 256^2}{248^2} \\\\T_2 = 372.95 \ Hz$