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Fynjy0 [20]
3 years ago
5

A spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. How fast is the radius of the balloon

increasing at the instant the radius is 40 centimeters
Physics
1 answer:
Nookie1986 [14]3 years ago
8 0

Answer:

<h2>0.245cm/min</h2>

Explanation:

The volume of the spherical balloon is expressed as V = 4/3πr³ where r is the radius of the spherical balloon. If the spherical balloon is inflated with gas at the rate of 500 cubic centimetres per minute then dV/dt = 500cm³.

Using chain rule to express dV/dt;

dV/dt = dV/dr*dr/dt

dr/dt is the rate at which the radius of the gallon is increasing.

From the formula, dV/dr = 3(4/3πr^3-1))

dV/dr = 4πr²

dV/dt = 4πr² *dr/dt

500 =  4πr² *dr/dt

If radius r = 40;

500 = 4π(40)² *dr/dt

500 = 6400π*dr/dt

dr/dt = 500/6400π

dr/dt = 5/64π

dr/dt  = 0.245cm/min

Hence, the radius of the balloon is increasing at the rate of 0.245cm/min

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Hope this helps!

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Why might some people not have believed Galileo's discoveries?
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They did not believed Galileo's discoveries because religiouse reasons the preast said that all the bible is true but Galileo despised it.
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First, you will investigate purely vertical motion. The kinematics equation for vertical motion (ignoring air resistance) is giv
Ivenika [448]

Answer:

2.85 s .

Explanation:

y(t) = y(0) + v₀t + 1/2 gt²

y(t) is vertical displacement , y(0) is initial position , v₀ is initial velocity and t is time required to make vertical displacement and g is acceleration due to gravity.

Here  y(0) is zero , v₀ = 14 m/s , g = 9.8 m s⁻² , y(t ) = 0 , as the pumpkin after time t comes back to its initial position, that is ground .

We shall take v₀ as negative as it is in upward direction and g as positive as it acts in downward direction

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8 0
3 years ago
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if 1.5kg block is pulled across a horizontal surface that has a coefficient of kintic friction of 0.60. if a block is pulled by
inysia [295]

Answer:

3.2N

Explanation:

Given parameters:

Mass of block  = 1.5kg

Coefficient of kinetic friction  = 0.6

Force of pull on block  = 12N

Unknown:

Net force on the block  = ?

Solution:

Frictional force is a force that opposes motion:

  Net force  = Force of pull  - Frictional force

 Frictional force  = umg

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   m is the mass

    g is the acceleration due to gravity

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4 0
3 years ago
A projectile is fired with an initial speed of 60.3 m/s at an angle of 34.2 above the horizontal on a long flat firing range.
amid [387]

Answer:

A.) H = 58.6 m

B.) T = 6.92 s

C.) 345.12 m

D.) V = 22.13 m/s

E.) Ø = 32.1 degree

Explanation:

Given that the

initial speed U = 60.3 m/s

Angle Ø = 34.2 degree

A.) At maximum height, final velocity V is equal to zero.

Using the third equation of motion under gravity.

V^2 = U sin Ø^2 - 2gH

Substitute for U and g. Where g = 9.8 m/s^2

0 = (60.3 sin 34.2)^2 - 2 × 9.8 × H

1148.78 = 19.6 H

H = 1148.78/19.6

H = 58.6 m

B.) To Determine the total time in the air, let us use the formula

V = UsinØ - gt

At maximum height, V = 0

t = UsinØ/g

Total time T = 2t

Therefore, T = 2UsinØ/g

T = (2 × 60.3 × sin 34.2)/9.8

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C.) To determine the total horizontal distance covered which is the range, we will use second equation of motion.

S = UcosØT - 1/2gt^2

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Substitute all the parameters into the formula

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D.) After 1.2 s firing,

V = UsinØ - gt

Where t = 1.2 s

Substitute into the formula

V = 60.3 × sin34.2 - 9.8 × 1.2

V = 33.89 - 11.76

V = 22.13 m/s

Therefore the speed of the projectile 1.20 s after firing is 22.13 m/s

E.) The direction will be determined by using the formula

t = VsinØ/ g

Cross multiply

VsinØ = gt

Make SinØ the subject of formula

SinØ = gt/V

SinØ = (9.8×1.2)/22.13

Sin Ø = 11.76/22.13

Sin Ø = 0.53

Ø = sin^-1( 0.53 )

Ø = 32.1 degree

3 0
3 years ago
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