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madreJ [45]
3 years ago
11

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

dius is 100 cm
Physics
1 answer:
Mrac [35]3 years ago
5 0
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
\frac{dA}{dt} = \frac{dA}{dr} \frac{dr}{dt}

The radius increases at the constant rate of 2 cm/min, therefore
\frac{dA}{dt} = 2 \frac{dA}{dr}=2*(8 \pi r) =16 \pi r

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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on the surface of theearth ,the weight of a boy is 400N but on a mountainpeak his weight is 360N,Calculatethe value of ''g' on t
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7 0
3 years ago
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9
11111nata11111 [884]

Answer:

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

m_{1} v_{1}=m_{2} v_{2}

\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }

\mathrm{v}_{1} \text { velocity before the collision }

\mathrm{v}_{2} \text { velocity after the collision }

\mathrm{m}_{1}=600 \mathrm{kg}

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\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}

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\mathrm{v}_{2}=\frac{3000}{900}

\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}

\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}

5 0
3 years ago
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