Answer:
The spillage of water
<em>PRECAUTION</em><em>:</em> Keep eureka can away from table edges and collect the displaced water in a measuring cylinder
<span>The momentum before the collision is equal to the momentum after the collision</span>
Answer:
a) 12.74 V
b) Two pairs of diode will work only half of the cycle
c) 8.11 V
d) 8.11 mA
Explanation:
The voltage after the transformer is relationated with the transformer relationshinp:
the peak voltage before the bridge rectifier is given by:
The diodes drop 0.7v, when we use a bridge rectifier only two diodes are working when the signal is positive and the other two when it's negative, so the peak voltage of the load is:
As we said before only two diodes will work at a time, because the signal is half positive and half negative,so two of them will work only half of the cycle.
The averague voltage on a full wave rectifier is given by:
Using Ohm's law:
Answer:
The speed of the wave remains the same
Explanation:
Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.
We observed that the speed, v is independent of the frequency of the wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.
<u>So, If you increase the frequency of oscillations, the speed of the wave remains the same.</u>
Answer:
σ = ±708 nC/m²
Q = ±177 nC
Explanation:
given data
Side of copper plate L = 50 cm
Electric field, E = 80 kN/C
solution
we get here Charge density,σ that is express as
σ = E x ε₀ ....................1
here ε₀ is Permittivity of free space that is 8.85 x 10⁻¹² C²/Nm²
so put value in eq1we get
σ = 80 x 10³ x 8.85 x 10⁻¹²
σ = 708 x 10⁻⁹ C/m²
σ = 708 nC/m²
and
now we get here total change on each faces
Q = σ A ...............2
Q = 708 x 10⁻⁹ x (0.50)²
Q = 177 nC
