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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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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0
3 years ago
Three charges, each of magnitude 10 nC, are at separate corners of a square of edge length 3 cm. The two charges at opposite cor
FinnZ [79.3K]

Answer:

The force exerted by three charges on the fourth is F_{resultant}=2.74\times10^{-5}\ \rm N

Explanation:

Given:

  • The magnitude of three identical charges, q=10\ \rm nC
  • Length of the edge of the square a=3 cm
  • Magnitude of fourth charge ,Q=3 nC

According to coulombs Law the force F between any two charge particles is given by

F=\dfrac{kQq}{r^2}

where r is the radial distance between them.

Since the force acting on the charge particle will be in different directions so according to triangle law of vector addition

F_{resultant}=\sqrt ((\dfrac{kQq}{L^2})^2+(\dfrac{kQq}{L^2} })^2)+\dfrac{kQq}{(\sqrt{2}L)^2}\\F_{resultant}=\dfrac{kQq}{L^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=\dfrac{9\times10^9\times10\times10^{-10}\times3\times10^{-9}}{0.03^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=2.74\times 10^{-5}\ \rm N

5 0
3 years ago
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