Answer:
0 - 60 mph = 0 - 26.8 m/s = 0 - 96.6 km/h; 0 - 100 km/h = 0 - 27.8 m/s = 0 - 62.1 mph.
Explanation:
Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
Answer:
90 J
Explanation:
W=fd
W=(75)(1.2)
W= 90 J
Answer:
The total number of oscillations made by the wave during the time of travel is 1.4 Oscillations. Strictly speaking, the number of complete oscillations is 1.
Explanation:
The required quantity is the number of complete oscillations made by the traveling wave. The amplitude time and frequency are not needed to calculate the number of oscillations as it is the ratio of the distance traveled to the wavelength( minimum distance that must be traveled to complete one oscillation) of the wave. So the total number of oscillations is 1.4 while the number of complete oscillations is 1 (strictly speaking). The detailed solution to this question can be found in the attachment below. Thank you!
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):

hence, the maximum speed is v_max = ((1/6)e^-1)a