During a climb UP the mountain, gravity does NO work on the climber.
Actually, it's more correct to say that gravity does NEGATIVE work
on him. The climber has to DO the positive work to haul himself up.
Work = (mass) x (gravity) x (height) .
For the guy in this problem:
Work = (67 kg) x (9.8 m/s²) x (3,500 meters)
= 2,298,100 joules.
If he eats no candy bars on the way, and completely depends on
his stored body fat for the energy, then he'll burn off
(2,298,100 joules) / (3.8 x 10⁷ joules/kg)
= 0.06 kg of fat.
That's only about 2.1 ounces. We KNOW he'll lose more weight than that,
climbing 11,000 feet. That's because climbing is pretty inefficient.
In addition to the potential energy you have to give your body weight,
you also have to expend energy breathing, digesting, metabolizing,
and sweating.
Answer:
when it hit the moving bat
Explanation:
force equals mass times acceleration which means the moving bat will add more force to the ball.
here three vectors are given as
now we will find the sum of all three
so its magnitude will be
So here this resultant will be same in all three cases as it will have same figure in all cases
1) G.mars= 9.8x(1/3)=3.26
Vi=10 m/s , T=12 sec
Dis. = Vi(t)+1/2(g)(t)^2
(10)(12)+1/2(3.26)(12)^2= 354.72
2)V=500 m/s
Hmax= v^/2(g) = 500^2/2(9.8) = 12755.1
Answer:
1.7 seconds
Explanation:
To clear the intersection, the total distance to be covered = 59.7 + 25 =84.7m
first we need to find the initial speed to just enter the intersection by using the third equation of motion
v^2 - u^2 = 2*a*s
45^2 - u^2 = 2 * -5.7 * 84.7
u^2 = 45^2 +965.58
u^2 = 2990.58
u = 54.7 m/s
Now for time we apply the first equation of motion
v-u =a * t
t = (v-u)/a = (45 - 54.7)/-5.7 = 1.7seconds