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3 years ago

Pressure and volume changes at a constant temperature can be calculated using

Physics

1 answer:

V125BC [204]3 years ago

3 0

Answer:

Option A, Boyle's law

Explanation:

The complete question is

Pressure and volume changes at a constant temperature can be calculated using

a. Boyle's law. c. Kelvin's law.

b. Charles's law. d. Dalton's law.

Solution

In Boyle’s law, the gas is assumed to be ideal gas and at constant temperature. With these two conditions fixed, Boyle’s established that volume of gas varies inversely with the absolute pressure.

The basic mathematical representation of this phenomenon is as follows -

$P \alpha \frac{1}{V}$

$P = \frac{k}{V} \\PV = k$

Where P is the pressure of ideal gas, V is the volume and k is the constant of proportionality.

Hence, option A is correct

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A cube-shaped piece of copper has sides of 4cm each and it's density is

iris [78.8K]

Answer:

$64 cm^3$

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

$\displaystyle \rho=\frac{m}{V}$

We have a cube-shaped piece of copper of 4 cm of side length. The volume of the piece is:

$V=(4\ cm)^3=64\ cm^3$

Surprisingly, no other magnitude is required, thus the answer is:

$\mathbf{64 cm^3}$

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2 years ago

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Nikitich [7]

Answer: vf = 51 m/s

d = 112 m

Explanation: Solution attached:

To find vf we use acceleration equation:

a = vf - vi / t

Derive to find vf

vf = at + vi

Substitute the values

vf = 3.5 m/s² ( 8.0 s) + 23 m/s

= 51 m/s

To solve for distance we use

d = (∆v)² / 2a

= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)

= (28 m/s)² / 7 m/s²

= 784 m/s / 7 m/s²

= 112 m

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3 years ago

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Stolb23 [73]

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3 years ago

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hello, I need help with a 25 question test! Which statement is an example of heat transfer by convection?Question 4 options:Jane

garri49 [273]

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1 year ago

To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the ho

cupoosta [38]

Answer:

xf = 5.68 × 10³ m

yf = 8.57 × 10³ m

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0 yi = 0

xf = ? yf = ?

vxi = vicosθ vyi = visinθ

ax = 0 ay = −9.8 m/s

now we solve x motion: that is

xf = xi + vxi × t + 0.5 × ax × t² ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ² ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²

yf = 8.57 × 10³ m

5 0

3 years ago