Earth has its own atmosphere. That is one reason all the water that has been on Earth has been recycled through the water cycle. It never leaves Earth’s atmosphere.
Answer:
If voltage is kept constant across the resistor itself, it' current will reduce. If the resistance is part of oscillator circuit, frequency response will change. If it is in series with capacitor or inductor, it will change the damping effect.
Explanation:
Power is the rate work done given by dividing work done by unit time. It is measured in watts equivalent to J/s.
In this case the force by the student is mg = 490 N (taking g as 9.8m/s²)
Work done is given by force × distance,
Therefore, Power =(force × distance)/ time, but velocity/speed =distance/time
Thus, Power = force × speed/velocity
= 490 N × 1.25
= 612.5 J/S (Watts)
Hence, power will be 612.5 Watts.
Answer:
To find the diameter of the wire, when the following are given:
Resistivity of the material (Rho), Current flowing in the conductor, I, Potential difference across the conductor ends, V, and length of the wire/conductor, L.
Using the ohm's law,
Resistance R = (rho*L)/A
R = V/I.
Crossectional area of the wire A = π*square of radius
Radius = sqrt(A/π)
Diameter = Radius/2 = [sqrt(A/π)]
Making A the subject of the formular
A = (rho* L* I)V.
From the result of A, Diameter can be determined using
Diameter = [sqrt(A/π)]/2. π is a constant with the value 22/7
Explanation:
Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation
Answer:
a.241.08 m/s b. 196 Hz c. 392 Hz
Explanation:
a. Determine the speed of waves within the wire.
The frequency of oscillation of the wave in the string, f = nv/2L where n = harmonic number, v = speed of wave in string, L = length of string = 1.23 m.
Since f = 588 Hz which is the 6 th harmonic, n = 6. So, making v subject of the formula, we have
v = 2Lf/n
substituting the values of the variables into v. we have
v = 2 × 1.23 m × 588Hz/6
v = 241.08 m/s
b. Determine the frequency at which the wire will vibrate with the first harmonic wave pattern.
The first harmonic is obtained from f when n = 1,
So, f = v/2L = 241.08 m/s ÷ 1.23m = 196 Hz
c. Determine the frequency at which the wire will vibrate with the second harmonic wave pattern.
The second harmonic f' = 2f = 2 × 196 Hz = 392 Hz