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3 years ago

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A spherical balloon is inflating with helium at a rate of 72 ft2/min . How fast is the balloon's radius increasing at the insta

nt the radius is 3 ft? How fast is the surface area increasing?

1 answer:

Yanka [14]3 years ago

7 0

The volume of the balloon is given by:

V = 4πr³/3

V = volume, r = radius

Differentiate both sides with respect to time t:

dV/dt = 4πr²(dr/dt)

Isolate dr/dt:

dr/dt = (dV/dt)/(4πr²)

Given values:

dV/dt = 72ft³/min

r = 3ft

Plug in and solve for dr/dt:

dr/dt = 72/(4π(3)²)

dr/dt = 0.64ft/min

The radius is increasing at a rate of 0.64ft/min

The surface area of the balloon is given by:

A = 4πr²

A = surface area, r = radius

Differentiate both sides with respect to time t:

dA/dt = 8πr(dr/dt)

Given values:

r = 3ft

dr/dt = 0.64ft/min

Plug in and solve for dA/dt:

dA/dt = 8π(3)(0.64)

dA/dt = 48.25ft²/min

The surface area is changing at a rate of 48.25ft²/min

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Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

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2 years ago

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A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at

Vesna [10]

Answer:

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0 So B=0

-16 B-32 A+16B=6 So A=-3/16

y=-3/16 cos 4t

Now to find the CF of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

$m^2+8m+16=0$

So $CF =(C_1+tC_2)e^{-2t}$

So the general equation

$Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t$

Given that t=0 Y=0 So

$C_1=\dfrac{3}{16}$

t=0 Y'=0 So

$C_2 =\dfrac{3}{8}$

$Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t$

The above equation is the general equation for motion.

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3 years ago

Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par

kondor19780726 [428]

Answer:

m1/m2 = 0.51

Explanation:

First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:

V = √F/u

This is the equation that describes the relation between speed of a pulse and a force exerted on it.

the value of "u" is:

u = m/L

Where m is the mass of the rod, and L the length.

Now, for the rod 1:

V1 = √F/u1 (1)

rod 2:

V2 = √F/u2 (2)

Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:

1.4V2 = √F/u1 (3)

Replacing (2) in (3):

1.4(√F/u2) = √F/u1 (4)

Now, let's solve the equation 4:

[1.4(√F/u2)]² = F/u1

1.96(F/u2) =F/u1

1.96F = F*u2/u1

1.96 = u2/u1 (5)

Now, replacing the expression of u into (5) we have the following:

1.96 = m2/L / m1/L

1.96 = m2/m1 (6)

But we need m1/m2 so:

1.96m1 = m2

m1/m2 = 1/1.96

m1/m2 = 0.51

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3 years ago