1 volt = 1 joule per coulomb.
Current doesn't actually pass 'through' a battery.
But if it did, then each coulomb would gain or lose 6 joules in traversing 6 volts, depending on its sign, and whether it climbed or fell.
Answer:
854.39 N
Explanation:
The formula for the fundamental frequency of a stretched string is given as,
f = 1/2L√(T/m)..................... Equation 1
Where f = fundamental frequency, L = Length of the wire, T = Tension, m = mass per unit length.
Given: f = 261.6 Hz, L = 0.6 m, m = (5.2×10⁻³/0.6) = 8.67×10⁻³ kg/m.
Substitute into equation 1
261.6 = 1/0.6√(T/8.67×10⁻³)
Making T the subject of the equation,
T = (261.6×0.6×2)²(8.67×10⁻³)
T =854.39 N
Hence the tension of the wire is 854.39 N.
To wear PPE equipment and read all the instructions before preforming the lab
Answer:
1. OK the
1. The horizontal axis of a osclloscope is generally TIME axis.
2. given T =25 micro second
That is = 25E-6 s
Then We f = (1/T) .
So f = 1/(25 x 10^-6) = 40000 Hz =
40 KHz.
3. First Band will be Yellow
Second Band will be Violet
Third Band will be Red
And
Fourth Band will be Gold
Thus, the value of the resistance = (47 x 100)plus or minus 5%
= (4700 plus or minua5%) ohm
the range of the acceptable value of the resistance will now be
= 4465 ohm to 4935 ohm
(4)
The level provided by the wall outlet = 120 x √2 = 169.68 V
The √2 is because the multimeter measures RMS voltage rather than peak voltage and
RMS Voltage is = √2 x peak voltage
Answer:
Wavelength of AM radio wave = 0.125 meter
Explanation:
Given:
Frequency of AM radio wave = 24 Hz
Velocity of AM radio wave = 3 m/s
Find:
Wavelength of AM radio wave
Computation:
Wavelength = Velocity / Frequency
Wavelength of AM radio wave = Velocity of AM radio wave / Frequency of AM radio wave
Wavelength of AM radio wave = 3 / 24
Wavelength of AM radio wave = 0.125 meter