negative acceleration- deceleration
Answer:
50 cubic cm
Explanation:
As the water displaces 50 cubic centimeters the amount of water that must leave the container for it to be able to be dropped into it is also 50 cubic as the water which was in its place must have gone somewhere.
The distance from point a and point is equal for travels going back and forth. They would just differ in time and speed because of the sea's speed that propels the vessel. The solutions as as follows:
Distance upstream = Distance downstream
99(t + 2) = (99 + 1)(t)
Solving for t,
t = 198 hours
That means that the distance is equal to:
Distance = 99 miles/hour * (198 + 2 hours)
Distance = 19,800 miles
Answer:
the intensity of the sound waves produced by one 60-w speaker at a distance of 1.0 m is 60 w/m²
Explanation:
Intensity of sound , I, is given as;
I = P/A
Where;
P is the power through an area = 60-w
A is the area = ?
A = 1.0m × 1.0m = 1.0 m²
I = 60-w / 1.0 m²
I = 60 w/m²
Therefore, the intensity of the sound waves produced by one 60-w speaker at a distance of 1.0 m is 60 w/m²
Answer:
.y₂ = 0.5704 m
, y2= 44.47 m,
Explanation:,
For this exercise we will use the kinematics relations
Ball
y₁ = y₀ + v₀₁ t
y₀ = 11.0 m
v₀₁ = 5.10 m / s
Pellet
y₂ = 0 + v₀₂ t - ½ g t²
V₀₂ = 39.0 m / s
At the meeting point the two bodies have the same height
y₁ = y₂
y₀ + v₀₁ t = v₀₂ t -1/2 g t²
11 + 5.1 t = 39 t - ½ 9.8 t²
4.9 t² - 33.9 t +11 = 0
.t2 - 6,918 + 2,245 = 0
t = [6,918 ±√ 6,918 2 - 4 2,245)] / 2
t = [6,918 ± 6.2356.] / 2
t₁ = 6.58 s
t₂ = 0.3412 s
Let's calculate the positions for each time
t₂ = 0.3412 s
y₂ = 39 t - ½ 9.8 t2
y₂ = 39 0.3112 - ½ 9.8 0.3412²
y₂ = 0.5704 m
.t1 = 6.58 s
.₂ = 39 6.58 - ½ 9.8 6.58 ^ 2
y2= 44.47 m