Answer:
Amplitude is decreased by a factor of
if intensity is decreased by a factor of 3.
Explanation:
Intensity of a sound wave is directly proportional to the square of its amplitude.
Therefore, if intensity is
and amplitude is
, then
, where,
is constant of proportionality.
Now, if intensity of sound wave is decreased by a factor of 3. So,
New intensity is, 

Plug in
for
. This gives,

Therefore, amplitude is decreased by a factor of
.
Answer:
<em>x < 4</em>
Explanation:
<u>Inequalities</u>
The inequalities relate expressions using signs different from the equal sign, like "<" or "<" among others.
They can be treated as a normal equation but in case of directional relationals, care must be taken when multiplying or dividing by negative numbers.
The inequality given in the question is
3x - 4 > 4x - 8
Subtracting 3x in both sides
3x - 4 -3x > 4x - 8 -3x
Simplifying
-4 > x - 8
Adding 8
-4 + 8 > x - 8 + 8
Simplifying
4 > x
The solution comes by flipping both sides and the inequality sign
x < 4
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
- ∑Fᵧ = maᵧ
- T₁ - m₂g = m₂aᵧ
Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
- T₁ = m₂aᵧ + m₂g
- T₁ = m₂(a + g)
We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
The normal force is equal to the x-component of the force of gravity.
- (F_n · μ_k) - m₁g sinΘ = m₁aₓ
- (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
- [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
- [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
- [2.539595871] - [-58.0962595] = 6aₓ
- 60.63585537 = 6aₓ
- aₓ = 10.1059759 m/s²
Now let's go back to this equation:
We have 3 known variables and we can solve for the tension force.
- T = 2(10.1059759 + 9.8)
- T = 2(19.9059759)
- T = 39.8119518 N
The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.
Jack------------ force of 92.5 n eastward-------Fjack(X)=92.5 n Fjack(Y)=0
<span>jill ------------------------------- force of 89.9 n northeast
Fjill(X)=cos45*89.9=63.57 n
</span>Fjill(Y)=sin45*89.9=63.57 n<span>
</span>jane -----------------------------force of 163 n southeast
Fjane(X)=cos45*163=115.26 n
Fjane(X)=-sin45*163=-115.26 n
Ftotal (X)=92.5+63.57+115.26=271.33 n
Ftotal (Y)=0+63.57-115.26=-51.69 n
Fotal=((271.33)^2+(-115.26)^2) ^0.5=294.80 n southeast
the magnitude of the net force the people exert on the donkey. is 294.80 n southeast