By constructing basic electric circuits, you will be able to measure current flow. With Ohm’s law in mind, what broad question a
2 answers:
Broad questions that we can answer by doing this experiment are <em><u>the effects of electric current</u></em>
<h3>Further explanation</h3>Electric current is the amount of electric charge that flows each unit of time
Electric current occurs due to the movement of electrons due to the difference in potential or voltage (from high potential to low potential) between two points
Electrons will flow through the conducting wire that functions as a conductor
Ohm's Law states that
<em>The potential difference is proportional to the electric current as long as the resistance is constant</em>
A simple electrical circuit consists of a voltage source (battery) and a lamp
Ampermeters to measure the strength of the current, must be installed in series with the load to be measured
By changing the voltage source, with constant resistance from a conductor, different current measurements will be obtained. The greater the voltage, the greater the resulting current.
<h3>Learn more</h3>electrons flow through the device
Keywords : basic electric circuits, Ohm’s law, experiment
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The distance (ft) traveled by the particle at time t (s) is
s(t) = 0.01 t⁴ - 0.02 t³
Part (a)
The velocity at time t is
v(t) = 0.04t³ - 0.06t² ft/s
Part (b)
After 1 s, the velocity is
v(1) = 0.04 - 0.06 = - 0.02 ft/s
Part (c)
When the particle is at rest, the velocity is zero. The time when this happens is given by
0.04t³ - 0.06t² = 0
t²(0.04t - 0.06) = 0
The graph shown below presents a clear picture of the motion.
Answer:
t = 0 (smaller value) or t = 1.5 s (larger value)
s(t) = 0.01 t⁴ - 0.02 t³
Part (a)
The velocity at time t is
v(t) = 0.04t³ - 0.06t² ft/s
Part (b)
After 1 s, the velocity is
v(1) = 0.04 - 0.06 = - 0.02 ft/s
Part (c)
When the particle is at rest, the velocity is zero. The time when this happens is given by
0.04t³ - 0.06t² = 0
t²(0.04t - 0.06) = 0
The graph shown below presents a clear picture of the motion.
Answer:
t = 0 (smaller value) or t = 1.5 s (larger value)
Answer:
600 Ω
Explanation:
when there is no load attached to the generator the circuit is open and zero current flows through the circuit, hence voltage drop across the internal resister is zero.
when 600Ω load is connected current starts flowing through the circuit and some voltage will drop across the internal resister.
voltage across the load resister is 1 V, so the current through it will be:
I=V/R
I=1/600 A= 1.67mA
voltage drop in the internal resister is:
V= input voltage - output voltage
V=2 V-1 V
⇒V=1 V
Now by using ohm's law
R=V/ I
R= (I= 1.67mA, as the resistors are connected in series)
⇒R=600 Ω
Hence Thevinin resistance of the generator is 600Ω.
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so = A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s