F the radius of a sphere is increasing at the constant rate of 2 cm/min, find the rate of change of its surface area when the ra
1 answer:
The surface area of a sphere of radius r is
A(r) = 4πr²
The rate of change of the surface area with respect to time is
The radius increases at the constant rate of 2 cm/min, therefore
When r = 100 cm, the rate of change of the surface area is
16π(100) cm²/min
= 1600π cm²/min
= 5026.5 cm²/min
Answer: 1600π or 5026.5 cm²/min
A(r) = 4πr²
The rate of change of the surface area with respect to time is
The radius increases at the constant rate of 2 cm/min, therefore
When r = 100 cm, the rate of change of the surface area is
16π(100) cm²/min
= 1600π cm²/min
= 5026.5 cm²/min
Answer: 1600π or 5026.5 cm²/min
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Explanation:
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Mass, m = 250 g = 0.25 kg
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mg=kx
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Answer:
D.) 1m/s
Explanation:
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where m is the mass of the pendulum, g = 10 m/s2 is the constant gravitational acceleration, h = 0.05 is the vertical it travels, v is the pendulum velocity at the bottom, which we are trying to solve for.
The m on both sides of the equation cancel out
so D is the correct answer