I believe it would be 2m/s.
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:

Answer:
24.57 revolutions
Explanation:
(a) If they do not slip on the pavement, then the angular acceleration is

(b) We can use the following equation of motion to find out the angle traveled by the wheel before coming to rest:

where v = 0 m/s is the final angular velocity of the wheel when it stops,
= 95rad/s is the initial angular velocity of the wheel,
is the deceleration of the wheel, and
is the angle swept in rad, which we care looking for:



As each revolution equals to 2π, the total revolution it makes before stop is
154.375 / 2π = 24.57 revolutions
Answer:
The tube should be held vertically and perpendicular to the ground.
Explanation:
Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:
Reasoning:
The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.
Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.
So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.
hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.