Answer:
See attached document
Explanation:
Entire process for deriving the asked expression dV across the bridge as function of dP is illustrated in the attachment below.
The document gives a step-by step process for arriving at the expression. However, manipulation of algebraic equations is skipped for the conciseness of the document.
It also gives the expression for the case when all resistors have different nominal values.
The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.
From the question given above, the following data were obtained:
Velocity of sound (v) = 343 m/s
Distance (x) = 8.42 m
Time (t) =?
We can obtain obtained the time as illustrated below:
v = 2x / t
343 = 2 × 8.42 / t
343 = 16.84 / t
Cross multiply
343 × t = 16.84
Divide both side by 343
t = 16.84/343
t = 0.05 s
Thus, the time between when the bat emits the sound and when it hears the echo is 0.05 s.
<h3>
How does a bat know how far away something is?</h3>
A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.
Learn more about time elapses between when the bat emits the sound :
<u>brainly.com/question/16931690</u>
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Correction question:
A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)
Answer:
I = M R^2 is the moment of inertia about a point that is a distance R from the center of mass (uniform distributed mass).
The moment of inertia about the center of a sphere is 2 / 5 M R^2.
By the parallel axis theorem the moment of inertia about a point on the rim of the sphere is I = 2/5 M R^2 + M R^2 = 7/5 M R^2
I = 7/5 * 20 kg * .2^2 m = 1.12 kg m^2
Answer:
a) Diffusion coefficient, D = 1.5 in/hr
b) Mean jump frequency, f = 0.0833 Hz
Explanation:
a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:
..........(1)
Where <r> = mean displacement
D = Diffusion coefficient
t = time = 12 hrs
sum of the squares of the distance divided by 100 is 36 in2.
<r>²= 36 in²
Substituting these values into equation (1) above
b) Mean jumping distance, <r> = 0.1 inches
Applying equation (1) again
Where D = 1.5 in/hr
The mean jump frequency, f = 1/t
f = 1/12
f = 0.0833 Hz
