The correct answer is rock cycle
Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency,
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f =
f =
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:
Temperature, The highness, and the time.
Hope this helps!
C=
Answer:
(A) -2940 J
(B) 392 J
(C) 212.33 N
Explanation:
mass of bear (m) = 25 kg
height of the pole (h) = 12 m
speed (v) = 5.6 m/s
acceleration due to gravity (g) = 9.8 m/s
(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)
height at the bottom = 0
= 25 x 9.8 x (0-12) = -2940 J
(B) kinetic energy of the Bear (KE) =
= = 392 J
(C) average frictional force =
- change in KE (ΔKE) = initial KE - final KE
- ΔKE = -
- when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes ΔKE = - 0 = 392 J
\frac{-(ΔKE+ΔU)}{h}[/tex] =
= = 212.33 N
To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as
Where,
I = Acoustic intensity in linear scale
= Hearing threshold
The value in decibels is 17dB, then
Using properties of logarithms we have,
Therefore the factor that the intensity of the sound was