Answer:
6858.5712 m/s
Explanation:
Given that:
Radius, r
R = 3.20 * 10^3.
Normal force = 0.5 * normal weight
Normal force = Fn ; Normal weight = Fg
Fn = 0.5Fg
Recall:
mv² / R = Fn + Fg
Fn = 0.5Fg
mv² / R = 0.5Fg + Fg
mv² /R = 1.5Fg
mv² = 1.5Fg * R
F = mg
mv² = 1.5* mg * R
v² = 1.5gR
v = sqrt(1.5gR)
V = sqrt(1.5 * 9.8 * 3.2 * 10^3)
V = sqrt(47.04^3)
V = 6858.5712 m/s
The formula for potential energy is
E(p) = mgh
(Mass x gravity x height)
Therefore energy = (5.3)(9.8)(6.6)
= 342.8 J
How did I get 9.8?
9.8 is the constant for gravity
Inertia
the awnswer is inertia b
Gravitational potential energy can be given by the equation
PE = mgh
where m is the mass,
g is the gravitational constant 9.81 or 10 depending on rounding
and h is the height
well weight is a force equiavlent to
W= m*g
so comparing that to the potential energy equation, divide the potential energy by the height and you will get weight in Newtons
Answer:
6.0 m/s
Explanation:
According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.
Therefore, we can write:
or
where:
m is the mass of the athlete
u is the initial speed of the athlete (at the bottom)
0 is the initial potential energy of the athlete (at the bottom)
v = 0.80 m/s is the final speed of the athlete (at the top)
is the acceleration due to gravity
h = 1.80 m is the final height of the athlete (at the top)
Solving the equation for u, we find the initial speed at which the athlete must jump:
