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:
When have passed 3.9[s], since James threw the ball.
Explanation:
First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.
We'll use the kinematics equations to find these two unknowns.
![y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\](https://tex.z-dn.net/?f=y%3Dy_%7B0%7D%20%2Bv_%7B0%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cy%3D%20elevation%20%5Bm%5D%5C%5Cy_%7B0%7D%3Dinitial%20height%20%5Bm%5D%5C%5Cv_%7B0%7D%3D%20initial%20velocity%20%5Bm%2Fs%5D%20%3D41.67%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20passed%20%5Bs%5D%5C%5Cg%3D%20gravity%20%5Bm%2Fs%5E2%5D%3D9.81%5Bm%2Fs%5E2%5D%5C%5CNow%20replacing%3A%5C%5Cy%3D0%2B41.67%20%2A%282%29-%5Cfrac%7B1%7D%7B2%7D%20%2A%289.81%29%2A%282%29%5E%7B2%7D%20%5C%5C%5C%5Cy%3D63.72%5Bm%5D%5C%5C)
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can find the velocity after 2 seconds.
![v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bo%7D%20%2Bg%2At%5C%5Creplacing%3A%5C%5Cv_%7Bf%7D%20%3D41.67-%289.81%29%2A%282%29%5C%5C%5C%5Cv_%7Bf%7D%3D22.05%5Bm%2Fs%5D)
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.
As we can see the equation is based on Time (t).
Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t
![y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%20%2Bv_%7Bo%7D%20%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%20%5C%5Cwhere%3A%5C%5Cv_%7Bo%7D%20%3D55.56%5Bm%2Fs%5D%20%3D%20initial%20velocity%5C%5Cy_%7Bo%7D%20%3D0%5Bm%5D%5C%5Cnow%20replacing%5C%5C63.72%20%2B22.05%20%2At-%284.905%29%2At%5E%7B2%7D%20%3D0%20%2B55.56%20%2At-%284.905%29%2At%5E%7B2%7D%20%5C%5C63.72%20%2B22.05%20%2At%20%3D0%20%2B55.56%20%2At%5C%5C63.72%20%3D%2033.51%2At%5C%5Ct%3D1.9%5Bs%5D)
Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.
Time = 2 + 1.9 = 3.9[s]
W=gm
where g - gravitation
m - mass
w - weight
as gravitation equals to zero, multiplying by 0 gives W=0
It is not possible to tell whether and object is heavy or light
Answer:
Therefore the correct statement is B.
Explanation:
In the interference and diffraction phenomena, the natural wave of electromagnetic radiation must be taken into account, the wave front that advances towards the slit can be considered as when it reaches it behaves like a series of wave emitters, each slightly out of phase from the previous one, following the Huygens principle that states that each point is compiled as a source of secondary waves.
The sum of all these waves results in the diffraction curve of the slit that has the shape
I = Io sin² θ /θ²
Where the angle is a function of the wavelength and the width of the slit.
From the above, the interference phenomenon can be treated as the sum of two diffraction phenomena displaced a distance equal to the separation of the slits (d)
Therefore the correct statement is B
Answer:
The tension in the rope is 262.88 N
Explanation:
Given:
Weight 
N
Length of rope 
m
Initial speed of ball 
For finding the tension in the rope,
First find the mass of rod,
( 
)
kg
Tension in the rope is,
N
Therefore, the tension in the rope is 262.88 N
