38.A student pushes a 0.15 kg box down against a spring doing 25 J of work on the spring. The student releases the box which lau
1 answer:
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Answer:
a) The object must have constant velocity.
d) The object must have zero acceleration.
Explanation:
We can solve the problem by using Newton's second law, which states that the net force acting on an object is equal to the product between mass and acceleration:
where
F is the net force
m is the mass of the object
a is the acceleration
In this problem, the net force on the object is zero:
F = 0
This means that the acceleration of the object is also zero, according to the previous equation:
a = 0
So statement (d) is correct. Moreover, acceleration is defined as the rate of change of velocity:
Which means that , so the velocity is constant. Therefore, statement (a) is also correct. The other two statements are false because:
b)The object must be at rest. --> false, the object can be moving at constant velocity, different from zero
c)The object must be at the origin. --> false, since the object can be in motion
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