Answer:
5.65 times
Explanation:
60 db sound is equal to 60 phons sound when frequency is kept at 1000Hz.
But when the frequency of sound is changed to 100 Hz , according to equal loudness curves , the loudness level on phon scale will be 35 phons.
A decrease of 10 phon on phon- scale makes sound 2 times less loud
Therefore a decrease of 25 phons will make loudness less intense by a factor equal to 2²°⁵ or 5.65 less intense . Therefore intensity at 100 Hz
must be increased 5.65 times so that its intensity matches intensity of 60 dB sound at 1000 Hz frequency.
The intensity of the sound wave is defined as the ratio between the power of the wave and the area through which the wave passes:
where
I is the intensity
P is the power
A is the area
If we use the data of the problem,
and
, we find the intensity of the sound wave:
Answer: left and applied force
Explanation:
Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
The distance traveled and the time it takes to travel are both going up at a steady rate. The distance goes from lap 1, lap 2, lap 3, lap 4. The time is increasing by a minute each time. When both the x and y scenario are increasing at a STEADY RATE the line will be linear. B displays a linear line.
