I think we will use the law of conservation of linear momentum;
M1V1 = M2V2
M1 = 4 kg (mass of the water balloon launcher)
V1=?
M2= 0.5 kg ( mass of the balloon)
V2 = 3 m/s
Therefore; 4 V1 = 0.5 × 3
4V1= 1.5
V1= 1.5/4
= 0.375 m/s
Answer:
The lowest possible frequency of sound for which this is possible is 1307.69 Hz
Explanation:
From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.
First, we will determine his distance from the second speaker using the Pythagorean theorem
l₂ = √(2.00²+5.00²)
l₂ = √4+25
l₂ = √29
l₂ = 5.39 m
Hence, the path difference is
ΔL = l₂ - l₁
ΔL = 5.39 m - 5.00 m
ΔL = 0.39 m
From the formula for destructive interference
ΔL = (n+1/2)λ
where n is any integer and λ is the wavelength
n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.
Then,
0.39 = (1+ 1/2)λ
0.39 = (3/2)λ
0.39 = 1.5λ
∴ λ = 0.39/1.5
λ = 0.26 m
From
v = fλ
f = v/λ
f = 340 / 0.26
f = 1307.69 Hz
Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.
Answer:
–77867 m/s/s.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:
Acceleration = (final velocity – Initial velocity) /time
a = (v – u) / t
With the above formula, we can obtain acceleration of the ball as follow:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
a = (v – u) / t
a = (–23.9 – 34.5) / 0.00075
a = –58.4 / 0.00075
a = –77867 m/s/s
Thus, the acceleration of the ball is –77867 m/s/s.
Answer:
Part a)
Part B)
Explanation:
Part A)
As we know that the point A lies on the top of the loop
so we will have by energy conservation
so the speed at point A is given as
Part B)
Now the force equation at point A is given as
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The object is moving, so at different times, it has different displacement. I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.
Displacement is the distance and the direction FROM (the position at the beginning) TO (the position at the end).
At the beginning ... time=0 ... the position is 1 meter.
At the end ... time=5 ... the position is zero.
The distance FROM the beginning TO the end is (zero - 1m) . That's <em>-1m </em>.
