The coil of wire in the center of the screen encompasses an area through which magnetic field lines pass, so there is a magnetic
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Answer:
a) The distance of spectator A to the player is 79.2 m
b) The distance of spectator B to the player is 43.9 m
c) The distance between the two spectators is 90.6 m
Explanation:
a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:
x = v * t
where:
x = position of the spectators
v = speed of sound
t = time
Then, the position for spectator A relative to the player is:
x = 343 m/s * 0.231 s = 79.2 m
b)For spectator B:
x = 343 m/s * 0.128 s
x = 43.9 m
The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.
c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:
(Distance AB)² = A² + B²
(Distance AB)² = (79.2 m)² + (43.9 m)²
Distance AB = 90. 6 m
Answer:
Explanation:
Value of spring constant K = weight / extension
=
=512 ft s⁻²
Frequency of oscillation of spring mass system
=
Putting the values we get
frequency
=[/tex]
= 4 .
Time period = 1 / frequency = .25 s
No of vibrations in 4π second
=
= 50 approx
When the weight is released 1 foot above equilibrium
PE stored = 1/2 K x² = .5 x 512 x 1 = 256 J
KE = 1/2 mv² = .5 x 32 x 4 = 64 J
Total energy = 320 J
Let x be the amplitude upto which spring stretches due to this energy
1/2 k x² = 320 ,
1/2 x 512 x² = 320
x = 1.12 ft approx