As per as my knowledge
The speed of a wave in a medium is affected by <u>d</u><u>e</u><u>n</u><u>s</u><u>i</u><u>t</u><u>y</u>,<u> </u><u>w</u><u>a</u><u>v</u><u>e</u><u>l</u><u>e</u><u>n</u><u>g</u><u>t</u><u>h</u> and <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u>:)
(Good luck on your test and mark me brainliest if this helps)
Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
400m in 32sec: (400/32)>12.5meters per second>
(12.5)(60)(60)(1/1000)=45km per hour
Constant speed would mean that the two forces are equivalent
At the bottom of the tank :
P = ρgH
P = (1000 kg/m³)(10 m/s²)(1 m)
P = 10000 N/m²
F = P • A
F = (10000 N/m²)(1 m²)
F = 10000 N
At the side of the tank :
Pav = ½ρgH
Pav = ½(1000 kg/m³)(10 m/s²)(1 m)
Pav = 5000 N/m²
F = P • A
F = (5000 N/m²)(1 m²)
F = 5000 N
