the path of an electron around the nucleus of an atom
Heat used by electric heater :
Q = m • c • ∆T
Q = (75 kg)(4200 J/kg°C)(43°C - 15°C)
Q = 8.82 × 10⁶ J
Cost of electrical energy :
Cost = (8.82 × 10⁶ J)/(3.6 × 10⁶ J) • ($ 0.15)
Cost = $ 0.3675
S=Vt
110=V(72)
110/72=V
V=1.527m/s
Answer:
23.5 mV
Explanation:
number of turn coil 'N' =22
radius 'r' =3.00 cm=>
0.03m
resistance = 1.00 Ω
B= 0.0100t + 0.0400t²
Time 't'= 4.60s
Note that Area'A' = πr²
The magnitude of induced EMF is given by,
lƩl =ΔφB/Δt = N (dB/dt)A
=N[d/dt (0.0100t + 0.0400 t²)A
=22(0.0100 + 0.0800(4.60))[π(0.03)²]
=0.0235
=23.5 mV
Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV
The block's velocity is determined as 10.03 m/s.
<u>Explanation:</u>
According to work energy theorem, the work done on an object is equal to the change in kinetic energy of the object.
So, work done = Kinetic energy
Thus, the velocity can be determined as
Velocity = 10.03 m/s.
So the block's velocity is determined as 10.03 m/s.
