that statement is true
a Third class lever applied when the effort place between the load and the fulcrum.
For example, in a forearm serve
Fulcrum : The elbow
Effort : The effort that putted by the biceps muscle
Load : The arm
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
Answer:
Cd(NO3)2 + Na2S --> CdS + 2 NaNO3
Explanation:
Answer:
a) 600 meters
b) between 0 and 10 seconds, and between 30 and 40 seconds.
c) the average of the magnitude of the velocity function is 15 m/s
Explanation:
a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:
![Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m](https://tex.z-dn.net/?f=Area%20%5C%2C%5C%2CTrapezoid%3D%28%5Cleft%5BB%2Bb%5D%5C%2C%28H%2F2%29%5C%5Cdisplacement%3D%20%5Cleft%5B%2840-0%29%2B%2830-10%29%5Cright%5D%20%5C%2C%2820%2F2%29%3D600%5C%2C%5C%2Cm)
b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.
Notice that between 10 and 30 seconds the velocity is constant (doesn't change) of magnitude 20 m/s, so in this section of the trip there is NO acceleration.
c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:

Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:
