The horizontal component of the initial velocity of the ball is 17.8cos(52°) = 10.6m/s and the vertical component is 17.8sin(52°) = 14.0m/s.
When the ball reaches its maximum height, its vertical component of velocity is 0 (at the highest point, the ball has no more upward velocity), so using the formula
v = u + at,
where v is the final velocity,
u is the initial velocity,
a is the acceleration due to gravity and t is the time taken to reach the highest point of the ball's trajectory. We can find t as u = 14.0m/s,
a = -9.8m/s² (negative due to gravity), and
v = 0:0 = 14.0 + (-9.8)t=> t = 1.43 seconds
The time taken for the ball to reach the ground from its highest point is equal to the time it takes for the ball to reach that highest point.
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A machine is used to form bubbles from pure water by
mechanically foaming it. The surface tension of water is 0:070 N
m-1. What is the gauge pressure inside bubbles of radius 10 m?
The gauge pressure inside the bubble is 14,000 N/m² or 14,000 Pa. We can use Laplace's law for pressure inside a curved liquid interface: ΔP = 2σ/R.
To find the gauge pressure inside bubbles, we can use the Laplace's law for pressure inside a curved liquid interface:
ΔP = 2σ/R
where ΔP is the pressure difference across the curved interface, σ is the surface tension of water, and R is the radius of the bubble.
Given:
Surface tension of water (σ) = 0.070 N/m
Radius of the bubble (R) = 10 μm = 10 × 10^(-6) m
Substituting the values into the equation, we have:
ΔP = 2σ/R
= 2 * 0.070 / (10 × 10^(-6))
= 14,000 N/m²
The gauge pressure is the difference between the absolute pressure inside the bubble and the atmospheric pressure. Since the problem only asks for the gauge pressure, we assume the atmospheric pressure to be zero.
Therefore, the gauge pressure inside the bubble is 14,000 N/m² or 14,000 Pa.
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