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:
a = 9.94 m/s²
Explanation:
given,
density at center= 1.6 x 10⁴ kg/m³
density at the surface = 2100 Kg/m³
volume mass density as function of distance

r is the radius of the spherical shell
dr is the thickness
volume of shell

mass of shell


now,

integrating both side



we know,




a = 9.94 m/s²
Answer:
the work that must be done to stop the hoop is 2.662 J
Explanation:
Given;
mass of the hoop, m = 110 kg
speed of the center mass, v = 0.22 m/s
The work that must be done to stop the hoop is equal to the change in the kinetic energy of the hoop;
W = ΔK.E
W = ¹/₂mv²
W = ¹/₂ x 110 x 0.22²
W = 2.662 J
Therefore, the work that must be done to stop the hoop is 2.662 J
The answer is 30 ... same as the Atomic number.