Answer:
76.78 km/h To calculate the average velocity for the total trip, you need to first determine the total distance traveled and the total time taken. First, let's calculate the total distance traveled. The trip consists of 2 legs. The 1st leg is 280 km and the 2nd leg is 210 km. So the total distance is 280 km + 210 km = 490 km. Now you need to calculate the total time taken. For this problem, there are 3 intervals that need to be accounted for. The travel time for the 1st leg, the duration of the rest stop in the middle, and the travel time for the 2nd leg. The travel time for both legs is calculated by dividing the distance traveled by the average speed. So for the first leg we have 280 km / (88 km / h) = 3.181818 h The 2nd leg is 210 km / (75 km/h) = 2.8 h The rest stop in hours is 24 min / (60 min/h) = 0.4 h The total time is 3.181818 h + 2.8 h + 0.4 h = 6.381818 h The average velocity is the distance divided by the time, giving: 490 km / (6.381818 h) = 76.78 km/h
Explanation:
Hope this helps!!
Answer:
Each of the joints represents a degree of freedom in the manipulator system and allows translation and rotary motion :) Hope this helps
Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T