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GaryK [48]
3 years ago
15

A spherical balloon is being inflated. Find the rate of increase of the surface area (S = 4Ï€r2) with respect to the radius r wh

en r is each of the following. (Answers in unit ft2/ft)
(a) 1 ft

(b) 3 ft

(c) 6 ft
Physics
1 answer:
erastovalidia [21]3 years ago
8 0

Answer:

A) 8π ft²/ft

B) 24π ft²/ft

C) 48π ft²/ft

Explanation:

Surface area of the spherical balloon is not clear here but it is supposed to be;

S = 4πr²

where:

r is the radius of the spherical balloon

So thus, the rate of change of the surface area of the spherical balloon by its radius will be:

dS/dr = 8πr

A) at r = 1ft;

dS/dr = 8 × π × 1

dS/dr = 8π ft²/ft

B) at r = 3 ft;

dS/dr = 8 × π × 3

dS/dr = 24π ft²/ft

C) at r = 6ft;

dS/dr = 8 × π × 6

dS/dr = 48π ft²/ft

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Answer:

x=2.4t+4.9t^2

Explanation:

This equation is one of the kinematic equations to solve for distance. The original equation is as follows:

X=Xo+Vt+1/2at^2

We know that the ball starts at rest meaning that its initial velocity and position is zero.

X=0+Vt+1/2at^2

Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.

X=Vt+4.9t^2

Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.

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X=2.4t+4.9t^2

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Describe the Pascal's principles​
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Explanation:

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In transistor emitter current is equal to which current?
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In transistor,
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2 years ago
3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0
2 years ago
How to find the coefficient of kinetic friction on an incline?
PolarNik [594]

Answer:

An object slides down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects.

Explanation:

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7 0
2 years ago
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