If the amplitude of a wave increases by a factor of 4, how is the intensity changed?
2 answers:
The intensity increases 16 times.
The relationship between intensity and amplitude is given below:
<span> I = constant
Since, A becomes 4 times, A' = 4A</span><span>
Thus the intensity becomes, I' = constant </span>× (<span>4A)</span>²<span>
</span>⇒<span> I' = 16A</span>²
Therefore, the intensity of the wave is increased by 16 times
The relationship between intensity and amplitude is given below:
<span> I = constant
Since, A becomes 4 times, A' = 4A</span><span>
Thus the intensity becomes, I' = constant </span>× (<span>4A)</span>²<span>
</span>⇒<span> I' = 16A</span>²
Therefore, the intensity of the wave is increased by 16 times
You might be interested in
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
The frictional force is 39.4 N
Explanation:
We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write
where
is the net force
m is the mass
a is the acceleration
Here we know that the box is moving with constant velocity, so its acceleration is zero:
This means that the net force is also zero:
The net force on the block is given by the applied force, forward, and the frictional force, backward:
where
is the applied force
is the frictional force
Therefore, solving for ,
Learn more about friction:
#LearnwithBrainly