Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
Gravitational potential energy<span> is </span>energy<span> an object possesses because of its position in a </span>gravitational<span> field. </span><span>The equation for </span>gravitational potential energy<span> is GPE = mgh.
GPE = 1200(1.6)(350) = 672000 J
Hope this answers the question. Have a nice day.</span>
Answer:
Watts=Volt*Amps
So A=570/120=4.75amps
If voltage drops to 110V We get A=570/110=5.(18...)amps
Given :
A bicycle racer rides from a starting marker to a turnaround marker at 10 m/s.
She then rides back along the same route from the turnaround marker to the starting marker at 25 m/s.
To Find :
Her average speed for the whole race.
Solution :
We know, average velocity is given by :

Therefore, average speed for the whole race is given by 17.5 m/s.