The answer is either answer B or answer C! Hope that helps
Answer:
32 km east
Explanation:
In order to find the final position of the balloon, we need to calculate its total displacement.
We have:
- Displacement towards west: 18 km
- Displacement towards east: 50 km
The two vectors are on the same line: this means that we can simply add them together (taking into account the proper direction) to obtain the total displacement.
Taking east as positive direction and west as negative direction, we have:

And since it is positive, it means that the final position of the balloon is 32 km east from its initial position.
Answer:
9.43*10^3 year
Explanation:
For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation
To start with, we use the formula
t(half) = In 2/k,
if we make k the subject of formula, we have
k = in 2/t(half), now we substitute for the values
k = in 2 / 5730
k = 1.21*10^-4 yr^-1
In(A/A•) = -kt, on rearranging, we find out that
t = -1/k * In(A/A•)
The next step is to substitite the values for each into the equation, giving us
t = -1/1.21*10^-4 * In(5.4/15.3)
t = -1/1.21*10^-4 * -1.1041
t = 0.943*10^4 year
Answer:
The angular velocity of the propeller is 2.22 rad/s.
Explanation:
The angular velocity (ω) of the propeller is:
Where:
θ: is the angular displacement = 10.6 revolutions
t: is the time = 30 s

Therefore, the angular velocity of the propeller is 2.22 rad/s.
I hope it helps you!
Answer:
the lowest possible frequency of the emitted tone is 404.79 Hz
Explanation:
Given the data in the question;
S₁ ← 5.50 m → L
↑
2.20 m
↓
S₂
We know that, the condition for destructive interference is;
Δr = ( 2m +
) × λ
where m = 0, 1, 2, 3 .......
Path difference between the two sound waves from the two speakers is;
Δr = √( 5.50² + 2.20² ) - 5.50
Δr = 5.92368 - 5.50
Δr = 0.42368 m
v = f × λ
f = ( 2m +
)v / Δr
m = 0, 1, 2, 3, ....
Now, for the lowest possible frequency, let m be 0
so
f = ( 0 +
)v / Δr
f =
(v) / Δr
we know that speed of sound in air v = 343 m/s
so we substitute
f =
(343) / 0.42368
f = 171.5 / 0.42368
f = 404.79 Hz
Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz