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
<span>Newton's law of gravitation is attractive, whereas Coulomb's law is attractive or repulsive. Both are proportional to the inverse square of distance.</span>
Several short trips taken from a cold start can use ...twice... as much fuel as a longer multi-purpose trip covering the same distance when the engine is warm.
In cold weather, properly designed gasoline aids in engine starting, while in hot weather, it helps prevent vapor lock. In order to meet the requirements of a modern engine, the fuel must have the volatility for which the engine's fuel system was built and an antiknock quality strong enough to prevent knock during routine operation.
During the intake phase, the air and fuel are combined before being introduced into the cylinder. The spark ignites the fuel-air mixture after the piston compresses it, resulting in combustion. During the power stroke, the piston is propelled by the expansion of the combustion gases.
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Answer:
ρ_c < ρ_gold the crown is not gold
Explanation:
For this exercise we use the translational equilibrium condition
∑ F = 0
in the air
T₁ - W = 0
T₁ = W
in water
T + B - W = 0
where T is the scale reading, B the liquid thrust and W the weight
B = W -T
B = 7.84 - 6.84
B = 1 N
thrust is
B = ρ g V_water
V_water = 
V_water = 
V_water = 1.02 10⁻⁴ m³
how the crown is totally submerged
V_c = V_water
V_c = 1.02 10⁻⁴ m³
now we can calculate the density of the crown
ρ_c = m_c / Vc
W = mg
m = W / g
we substitute
ρ_c = 
ρ_c = 
ρ_c = 0.784 10⁴ kg / m³
ρ_c = 7.84 10³ kg / m³
We can see that this density is less than the density of gold
ρ_c < ρ_gold
therefore the crown is not gold
