Answer:
2.029×10^-18 J
Explanation:
E=hv
so
E=(3.06×10^15)*(6.63×10^-34)
E=2.029×10^-18 J
Answer: C
Both Technicians A and B
Explanation:
Only a DOT-approved flasher unit should be used for turn signals. And a parallel (variable-load) flasher will function for turn signal usage, although it will not warn the driver if a bulb burns out.
Answer:
6.96 s
Explanation:
<u>Given:</u>
- u = initial speed of the automobile = 0 m/s
- a = constant acceleration of the automobile =

- v = constant speed of the truck = 8.7 m/s
<u>Assume:</u>
- t = time instant at which the automobile overtakes the truck.
At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.
Nearly equal the output work is greater than the input work because of friction.All machines use some amount of input work to overcome friction.The only way to increase the work output is to increase the work you put into the machine.You cannot get more work out of a machine than you put into it.