Given Information:
Angular velocity = ω = 4 rad/s
Angular acceleration = α = 5 rad/s²
Center deceleration = a₀ = 2 m/s
Required Information:
Acceleration of point A at this instant = ?
Answer:
Acceleration of point A at this instant = 5.94 m/s²
Explanation:
Refer to the attached diagram of the question,
The acceleration of point A is given by
a = a₀ + rα - rω²
Where r is the radial distance between the center and point A, a₀ is the deceleration of center, α is the angular acceleration and ω is the angular velocity.
a = -2i + 0.3j*5k - 0.3j*4²
a = -2i + 1.5(j*k) - 0.3j*16
a = -2i + 1.5(-i) - 4.8j
a = -2i - 1.5i - 4.8j
a = -3.5i - 4.8j
The magnitude of acceleration vector is
a = √(-3.5)² + (-4.8)²
a = √35.29
a = 5.94 m/s²
Therefore, the acceleration of point A is 5.94 m/s²
The angle is given by
θ = tan⁻¹(y/x)
θ = tan⁻¹(-4.8/-3.5)
θ = 53.9°
Answer:
The reactances vary with frequency, with large XL at high frequencies and large Xc at low frequencies, as we have seen in three previous examples. At some intermediate frequency fo, the reactances will be the same and will cancel, giving Z = R; this is a minimum value for impedance and a maximum value for Irms results. We can get an expression for fo by taking
XL=Xc
Substituting the definitions of XL and XC,
2
foL=1/2foC
Solving this expression for fo yields
fo=1/2
where fo is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if it were not driven by the voltage source. In fo, the effects of the inductor and capacitor are canceled, so that Z = R and Irms is a maximum.
Explanation:
Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined as a forced oscillation, in this case, forced by the voltage source, at the natural frequency of the system. The receiver on a radio is an RLC circuit that oscillates best at its {f} 0. A variable capacitor is often used to adjust fo to receive a desired frequency and reject others is a graph of current versus frequency, illustrating a resonant peak at Irms at fo. The two arcs are for two dissimilar circuits, which vary only in the amount of resistance in them. The peak is lower and wider for the highest resistance circuit. Thus, the circuit of higher resistance does not resonate as strongly and would not be as selective in a radio receiver, for example.
A current versus frequency graph for two RLC series circuits that differ only in the amount of resistance. Both have resonance at fo, but for the highest resistance it is lower and wider. The conductive AC voltage source has a fixed amplitude Vo.
Answer:
2102.1 m
Explanation:
Temperature at the equator = 0⁰
Radius of the earth = 6.37x10⁶
Required:
We how to find out what the clearance between tape and ground would be if temperature increases to 30 degrees.
Final temperature = ∆T = 303-273 = 30
S = 11x10^-6
The clearance R = Ro*S*∆T
=6.37x10⁶x 11x10^-6x30
= 2102.1m
Or 2.102 kilometers
Thank you
Answer:
False......................
Any Voltage used on board a ship if less than 1kV (1000 V) then it is called as LV (Low Voltage) system and any voltage above 1kV is termed as High Voltage. Typical Marine HV systems operate usually at 3.3kV or 6.6kV.
Explanation:
trust
