Dave and Alex push on opposite ends of a car that has a mass of 875 kg. Dave pushes the car to the right with a force of 250 N,
1 answer:
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Answer:
hello your question has some missing parts below is the complete question
and the missing diagram
The two speakers emit sound that is 180° out of phase and of a single frequency,ƒ, Find the lowest two frequencies that produce a maximum sound intensity at the positions of Moe and Curly.
answer : 1316.2 hertz
Explanation:
The frequency that produce the maximum sound intensity can be calculated using the relation below
dsin ∅ = n <em>A</em>
where <em>A = </em>dsin ∅ / n when n = 1 . d = 0.800
<em>A</em> = 0.800 * ( 1 / 3.162 )
<em>A</em> = 0.253 m
speed of sound = 333 m/s
frequency = speed /<em> A</em>
<em>= </em>333 / 0.253 = 1316.2 hertz
Answer:
A)
B)
Explanation:
A) In this situation we are talking about a car moving only in the X- axis, hence the velocity of the car is:
Where the unit vectors ,
and
represent the components
,
and
in the cartesian plane.
In this sense, each unit vector is defined to have a magnitude of exactly one (1).
B) Velocity is defined as the variation of position in time, if this car is moving only along the x direction we will have:
Clearing the position:
Answer:
Minimum capacitance = 200 μF
Explanation:
From image B attached, we can calculate the current flowing through the capacitors.
Thus;
Since V=IR; I = V/R = 5/500 = 0.01 A
Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V
So minimum capacitance will be determined from;
i(t) = C(dv/dt)
So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]
C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF
