= 27.777
Explanation:
A kilometer has 1,000 meters, and an hour has 3,600 seconds, so 100 kilometers per hour is: 100 x 1,000 / 3,600 = 27.777... m/s.
Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
If it's volume changes when you move it to the new container it would be a solid
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:
a) y₂ = 49.1 m
, t = 1.02 s
, b) y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
² =
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
y-yo =
=
- g t
t =
/ g
t = 10 / 9.8
t = 1.02 s
b) the maximum height
y- 44.0 =
² / 2 g
y - 44.0 = 5.1
y = 5.1 +44.0
y = 49.1 m
The time is the same because it does not depend on the initial height
t = 1.02 s