Answer:
Induced EMF is 2 x 10⁻³ volts
Explanation:
B = strength of the magnetic field aligning with the loop's axis = 1 T
= area change rate of the loop = 20 cm²/s = 20 x 10⁻⁴ m²
θ = the angle formed by the magnetic field and area vector = 0
E = the induced EMF across the loop
EMF can be calculated using the formula
E = B
E = (1) (20 x 10⁻⁴ )
E = 2 x 10⁻³ volts
E = 2 mV
Answer:
Every option provided is accurate
Explanation:
The electrical power dissipated by a single resistor linked to a battery can be expressed as:
where
V signifies the voltage
I denotes the current
R represents the resistance
Now, let's evaluate each scenario:
A) When the voltage is doubled (V'=2V) while the current is halved (I'=I/2), the resulting power dissipation turns out to be:
--> the power remains the same
B) When the voltage is increased to double (V'=2V) and the resistance quadruples (R'=4R), the new power dissipation becomes:
--> the power is unchanged
C) If the current is doubled (I'=2I) while the resistance diminishes to one-fourth (R'=R/4), the new power dissipation is:
--> the power is unchanged
Answer:
a) ∆x∆v = 5.78*10^-5
∆v = 1157.08 m/s
b) 4.32*10^{-11}
Explanation:
This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:
Where h represents Planck’s constant (6.62*10^-34 J s).
Assuming that the electron's mass remains the same, we proceed as follows:
Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:
If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity: