We use the kinematic equations,
(A)
(B)
Here, u is initial velocity, v is final velocity, a is acceleration and t is time.
Given,
,
and
.
Substituting these values in equation (B), we get
.
Therefore from equation (A),

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.
<h3>
Angular velocity of the tire</h3>
The angular velocity of the tire is the rate of change of angular displacement of the tire with time.
The magnitude of the angular velocity of the tire is calculated as follows;
ω = 2πN
where;
- N is the number of revolutions per second
ω = 2π x (5.25 / 3)
ω = 11 rad/s
<h3>Tangential velocity of the tire</h3>
The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.
The magnitude of the tangential velocity is caculated as follows;
v = ωr
where;
- r is the radius of the car's tire
v = 11r m/s
Learn more about tangential velocity here: brainly.com/question/25780931
Answer:
Speed is a measure of how quickly an object moves from one place to another. It is equal to the distance traveled divided by the time.
Answer:
The magnitude of the force net is an acting object multiplied by acceleration of the object
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