Uh so I'm no master at this subject, but all stuffs accelerate at 9.8 m/s squared. So you multiply the 9.8 and the 0.20 it's given for reasons unknown other than that's what I see in my notes... and that gives you 1.96 m/s squared.
As for B, I have no idea. I think you may multiply the 1.96 by 4. Tell me your thoughts and maybe we can work it out together
(a) Let's convert the final speed of the car in m/s:

The kinetic energy of the car at t=19 s is

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:

But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into

(c) The instantaneous power is given by

where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:

And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
Answer:
Yes the body will receive a dangerous shock in both cases.
Explanation:
Different parts of the body has different resistance. skin has the high resistance as compared to other organs of the body.
Dry skin has high resistance than wet skin this is because water is relatively good conductor of electricity, it adds parallel path to the current flow and hence reduces skin resistance.
Dry hands body has approximately 500 kΩ resistance and if 120 V electricity supply current received will be:
I = V/R= 120/ 500*10^3
I= 0.24 mA
Even the current seems is much lower than the safe zone but this is the case in case of DC voltage in case of AC voltage the body will receive a shock this is because the skin pass more current when the voltage is changing i.e. AC.
Similarly for wet hands body resistance is 1 kΩ. so the current through the body seems to be:
I = 120 / 1000
I = 12 mA
The current is higher than safe zone so the body will receive a dangerous shock.
Except when necessary for takeoff and landing, <span>the minimum safe altitude required for a pilot to operate an aircraft over other than congested area is an altitude of 1000 ft above the highest obstacle within a 2000 ft horizontal radius of the aircraft.
It is also good to know that apart from taking off and landing, the aircraft must not operate at a distance less than 500 ft from any person, vessel, structure or vehicle.</span>
Answer:
18.60 m/s
Explanation:
Original momentum = mv = 4000 with m = 115
after collision m = 115 + 100 = 215 kg
but the total momentum is still the same (conserved)
4000 = 215 v shows v = 18.60 m/s