Ik what it is hit me up for it
Answer:
80 meters high
Explanation:
The velocity of the balloon would be g*t (I won't calculate, but will us this later)
We know that the kinetic energy at the bottom equals the potential at the top.
KE = PE
1/2 * m * v^2 = m * g * h
1/2 * m * (g * t)^2 = m * g * h (substitution)
1/2 * m * g^2 * t^2 = m * g * h
1/2 * g * t^2 = h (simplification by dividing the commons between both sides)
h = 1/2 * 9.81 * 4^2
h = 78.48 m (roughly 80 m)
Answer:
a = -2.4 m/s²
Explanation:
Given,
The initial speed of the bus, u = 24 m/s
The final speed of bus, v = 12 m/s
Time taken to reach final speed is, t = 5.0 s
The acceleration of the body is given by the change in velocity by time
a = (v - u) / t
= (12 - 24) / 5
= -2.4 m/s²
The negative sign in the acceleration indicates that the bus is decelerating.
Therefore, the acceleration of the bus is, a = -2.4 m/s²
Answer:
maximum speed of the bananas is 18.8183 m/s
Explanation:
Given data
amplitude A = 23.195 cm
spring constant K = 15.2676 N/m
mass of the bananas m = 56.9816 kg
to find out
maximum speed of the bananas
solution
we know that radial oscillation frequency formula that is = √(K/A)
radial oscillation frequency = √(15.2676/23.195)
radial oscillation frequency is 0.8113125 rad/s
so maximum speed of the bananas = radial oscillation frequency × amplitude
maximum speed of the bananas = 0.8113125 × 23.195
maximum speed of the bananas is 18.8183 m/s
