How much current will a 500 W vacuum cleaner draw if it has a resistance of 30Ω?
1 answer:
Answer:
Electric current, I = 4.08 A
Explanation:
We have,
Power of vacuum cleaner is 500 W
Resistance of vacuum cleaner is 30 ohms
It is required to find the current flowing through the vacuum cleaner. It can be calculated using the definition of power delivered. Its formula is given by :
I is electric current
So, the electric current flowing through the vacuum cleaner is 4.08 A.
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Answer:
a) the magnitude of the force is
F= Q() and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E =
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = , where p = q × s
E(r) = where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q()
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude
ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
Answer:
Wavelength of radio is wave is 3 m
Explanation:
Wavelength of radio is wave is
where
wavelength is