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3 years ago

A balloon is filled with helium at a pressure of 2.4 x 105 Pa. The balloon is at a

1 answer:

5 0

1)
p = 2.4 * 10^5 Pa
T = 18° C + 273.15 = 291.15 k
r = 0.25 m => V = [4/3]π(r^3) = [4/3]π(0.25m)^3 = 0.06545 m^3 = 65.45 L

Use ideal gas equation: pV = nRT => n = pV / RT = [2.4*10^5 Pa * 0.06545 m^3] / [8.31 J/k*mol * 291.15k] = 6.492 mol

Avogadro number = 1 mol = 6.022 * 10^23 atoms

Number of atoms = 6.492 mol * 6.022 *10^23 atom/mol = 39.097 * 10^23 atoms = 3.91 * 10^24 atoms

2) Double atoms => double volume

V2 / V1 = r2 ^3 / r1/3

2 = r2 ^3 / r1 ^3 => r2 ^3 = 2* r1 ^3

r2 = [∛2]r1

The factor is ∛2

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A concave lens has a focal length of -31 {cm}. Find the image distance and magnification that results when an object is placed 2

kolezko [41]

Answer:

The value is $v = 440 \ cm$

and

$m = 16$

Explanation:

From the question we are told that

The image distance is $v = 29 \ cm$

The focal length is $f = -31 \ cm$

From the lens equation we have that

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

=> $\frac{1}{-31} = \frac{1}{v} - \frac{1}{29}$

=> $v = 450 \ cm$

Generally the magnification is

$m = \frac{v}{u}$

=> $m = \frac{ 450}{29}$

=> $m = 16$

7 0

3 years ago

Calculate the cross-sectional area of a cylinder of radius 25 mm

bezimeni [28]

Answer:

12.5

Explanation:

may i be marked brainliest?

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3 years ago

Does an object need a force to start moving

nadya68 [22]

Newton's first law actually states that a force isn't needed to keep an object moving, so no.

8 0

3 years ago

Which equation is used to calculate the electric potential in an electric field from a point charge?

scZoUnD [109]

Answer:

The electric potential (voltage) $V$ produced by a point charge $Q$ , at any point in space, is given by the following equation:

$V=k\frac{Q}{r}$

Where:

$k=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$r$ is the distance

The result is a scalar quantity, is defined as the electric potential energy per unit of charge and determines the electric influence exerted by the charge on that point of space.

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3 years ago

An object is dropped out of an airplane that is moving horizontally at 350 m/s and is 22,000 m above the ground. Ignoring fricti

lilavasa [31]

Answer:

Vx = 350m/s

Vy = 656.99 m/s

Explanation:

at the time of drop, the horizontal velocity of the object is same as of plane.

Since friction is ignored, horizontal velocity remains unchanged.

Vx = 350m/s

but the Vy increases due to gravitational acceleration

V^2 - U^2 = 2as

V: FINAL VELOCITY

U: INITIAL VELOCITY

U = 0 m/s

$v=\sqrt{2as}= \sqrt{2(9.81)(22000)} = 656.99 m/s$

8 0

3 years ago