Answer:
The height of the bridge is 78.4 m.
Explanation:
Given;
time of the stone motion off the bridge, t = 4.0 s
acceleration due to gravity, g = 9.8 m/s²
The height of the bridge is given by;
h = ut + ¹/₂gt²
where;
u is the initial velocity of the stone, u = 0
h = ¹/₂gt²
h = ¹/₂(9.8)(4)²
h = 78.4 m
Therefore, the height of the bridge is 78.4 m.
Not sure... need help with it
Work done = 0.5*m*[(v2)^2 - (v1)^2]
where m is mass,
v2 and v1 are the velocities.
Given that m = 1.50 x 10^3 kg, v2 = -15 m/s (decelerates), v1 = 25 kg,
Work done = 0.5 * 1.50 x 10^3 * ((-15)^2 - 25^2) = 3 x 10^5 joules
Just ignore the negative value for the final result because work is a scalar quantity.
The radius, r, of the child from the center of the wheel is
r = 1.3 m
The wheel makes one revolution in 4.2 s. Its angular velocity is
ω = (2π rad)/(4.2 s) = 1.496 rad/s
The linear speed of the child is the tangential velocity, given by
v = rω
= (1.3 m)*(1.496 rad/s)
= 1.945 m/s
Answer: 1.95 m/s (nearest hundredth)
