Answer:
(a) 3.807 s
(b) 145.581 m
Explanation:
Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.
The distance traveled by car after Δt (seconds) at
speed is
![s_c = \Delta t v_c = 23\Delta t](https://tex.z-dn.net/?f=s_c%20%3D%20%5CDelta%20t%20v_c%20%3D%2023%5CDelta%20t)
The distance traveled by the motorcycle after Δt (seconds) at
speed and acceleration of a = 8 m/s2 is
![s_m = \Delta t v_m + a\Delta t^2/2](https://tex.z-dn.net/?f=s_m%20%3D%20%5CDelta%20t%20v_m%20%2B%20a%5CDelta%20t%5E2%2F2)
![s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2](https://tex.z-dn.net/?f=s_m%20%3D%2023%5CDelta%20t%20%2B%208%5CDelta%20t%5E2%2F2%20%3D%2023%20%5CDelta%20t%20%2B%204%20%5CDelta%20t%5E2)
We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:
![s_m = s_c + 58](https://tex.z-dn.net/?f=%20s_m%20%3D%20s_c%20%2B%2058)
![23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58](https://tex.z-dn.net/?f=23%20%5CDelta%20t%20%2B%204%20%5CDelta%20t%5E2%20%3D%2023%5CDelta%20t%20%2B%2058)
![4 \Delta t^2 = 58](https://tex.z-dn.net/?f=4%20%5CDelta%20t%5E2%20%3D%2058)
![\Delta t^2 = 14.5](https://tex.z-dn.net/?f=%20%5CDelta%20t%5E2%20%3D%2014.5)
![\Delta t = \sqrt{14.5} = 3.807s](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%20%5Csqrt%7B14.5%7D%20%3D%203.807s)
(b)
![s_m = 23 \Delta t + 4 \Delta t^2](https://tex.z-dn.net/?f=s_m%20%3D%2023%20%5CDelta%20t%20%2B%204%20%5CDelta%20t%5E2)
![s_m = 23*3.807 + 58 = 145.581 m](https://tex.z-dn.net/?f=s_m%20%3D%2023%2A3.807%20%2B%2058%20%3D%20145.581%20m)
In Electrostatics the electrical force between Two charged objects is inversely Related to the distance of separation between the two objects .
Answer: (A) At terminal velocity ...
Explanation:
S = frequency * wavelength
s = 100 * 0.30
s = 30 Hz
hope this helps :)
Answer:
490N
Explanation:
According Newton's second law!
\sum Force = mass × acceleration
Fm - Ff = ma
Fm is the moving force
Ff s the frictional force = 100N
mass = 65kg
acceleration = 6m/s²
Required
Moving force Fm
Substitute the given force into thr expression and get Fm
Fm -100 = 65(6)
Fm -100 = 390
Fm = 390+100
Fm = 490N
Hence the force that will cause two cart to move is 490N