Answer:
Distance, d = 112.5 meters
Explanation:
Initially, the bicyclist is at rest, u = 0
Final speed of the bicyclist, v = 30 m/s
Acceleration of the bicycle, ![a=4\ m/s^2](https://tex.z-dn.net/?f=a%3D4%5C%20m%2Fs%5E2)
Let s is the distance travelled by the bicyclist. The third equation of motion is given as :
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
![s=\dfrac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
![s=\dfrac{(30)^2}{2\times 4}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7B%2830%29%5E2%7D%7B2%5Ctimes%204%7D)
s = 112.5 meters
So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.
Answer:
Option D) 4A
Explanation:
As the cycle of the wave passes by, the amplitude gives the longest journey when the spot travels from the undistributed position. During each cycle the spot travels "Four times" .
Considering one of this cycle, if it begins to travel from it's undistributed position , there would be four movements i.e
* Upward movement through distance A
*Downward movement through distance A
*Downward again through distance A
*Upward through distance A.
Then it would travel back to its undistributed position held
Answer:
796.18 Hz
Explanation:
Applying,
Maximum velocity = Amplitude×Angular velocity
Therefore,
V' = A(2πf)............... Equation 1
Where V' = maximum velocity of the eardrum, A = Amplitude of vibration of the eardrum, f = frequency of the eardrum vibration, π = pie
make f the subject of the equation
f = V'/2πA................ Equation 2
From the question,
Given: V' = 3.6×10⁻³ m/s, A' = 7.2×10⁻⁷ m,
Constant: 3.14.
Substitute these values into equation 2
f = 3.6×10⁻³/( 7.2×10⁻⁷×2×3.14)
f = 796.18 Hz