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photoshop1234 [79]
3 years ago
7

A tank having a volume of 0.12 m3 contains helium gas at 150 atm. How many balloons can the tank blow up, if each filled balloon

is a sphere of radius 0.16 m at an absolute pressure of 1.2 atm?
Physics
1 answer:
Ratling [72]3 years ago
8 0

Answer:

877

Explanation:

P1 = 150 atm

V1 = 0.12 m^3

P2 = 1.2 atm

V2 = ?

radius of each balloon, r = 0.16 m

Volume of each balloon,

V_{2}=\frac{4}{3}\pi r^{3}

V_{2}=\frac{4}{3}\pi 0.16^{3}

V2 = 0.0171 m^3

Let the number of balloons be N.

So,

P1 x V1 = N x P2 x V2

150 x 0.12 = N x 1.2 x 0.0171

N = 877.19

So, the number of balloons be 877.

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A scientific hypothesis must be tetable so it can become a scientific theory.

Explanation: I think

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3 years ago
A person weighing 645 N climbs up a ladder to a height of 4.55 m what is the increase in gravitational energy
Lemur [1.5K]

Answer: 2934.75 Joules

Explanation:

Potential energy can be defined as energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

<em>P.E = mgh</em>

Where P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per second square.

h represents the height measured in meters.

Given the following data;

Weight =645

Height = 4.55

<em>P.E = mgh</em>

But we know that weight = mg = 645N

Substituting into the equation, we have;

<em>P.E = 645 • 4.55</em>

<em>P.E = 2934.75J</em>

Potential energy, P.E = 2934.75 Joules.

3 0
3 years ago
A spring has a force constant of 310.0 N/m. (a) Determine the potential energy stored in the spring when the spring is stretched
skad [1K]

Answer:

(a) 0.2618 J

(b)  0.1558 J

(c) 0 J

Explanation:

from Hook's Law,

The energy stored in a stretched spring = 1/2ke²

Ep = 1/2ke² ......................... Equation 1

Where k = spring constant, e = extension, E p = potential energy stored in the spring.

(a) When The spring is stretched to 4.11 cm,

Given: k = 310 N/m, e = 4.11 cm = 0.0411 m

Substituting these values into equation 1

Ep = 1/2(310)(0.0411)²

Ep = 155(0.0016892)

Ep =155×0.0016892

Ep = 0.2618 J.

(b) When the spring is stretched 3.17 cm

e = 3.17 cm = 0.0317 m.

Ep = 1/2(310)(0.0317)²

Ep = 155(0.0317)²

Ep = 155(0.0010049)

Ep = 0.155758 J

Ep ≈ 0.1558 J.

(c) When the spring is unstretched,

e = 0 m, k = 310 N/m

Ep = 1/2(310)(0)²

Ep = 0 J.

6 0
3 years ago
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