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3 years ago

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State the Boyle's Law. If the graph for Boyle's law is straight line through the origin then state the quantity taken along: a)

X-axis , b)Y-axis.(NOTE: Please clearly mention the quantities appropriately by mentioning the axes)

1 answer:

astraxan [27]3 years ago

7 0

Answer:

(a) X- axis represent (1/P) inverse of pressure

(b) Y-axis represent (V) volume

Explanation:

Boyle's Law States that at constant temperature, the volume of a fixed mass of gas is inversly proportional to its pressure. It can be expressed mathematically as

V α 1/P

VPα1

removing the proportionality sign and introducing a constant

VP = k

V₁P₁ = V₂P₂

Where V₁ and V₂ are initial and final volume respectively, P₁ and P₂ are initial and final pressure respectively.

If the graph for Boyle's law is straight line through the origin,

(a) X-axis represent (1/P) inverse of pressure

(b) Y- axis represent (V) volume.

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Answer:

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N μ = frictional of incline on object where N is the normal force

N = m g cos theta force perpendicular to incline

m g sin theta = N μ = μ m g cos theta

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A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

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4 years ago

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3 years ago

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a net force of 100 newton’s is a moving mass with an acceleration of 5 m/s. what is the mass of the object?

Jlenok [28]

F = ma

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100 / 5 = m

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3 years ago

A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the

Elina [12.6K]

Answer:

g ±Δg = (9.8 ± 0.2) m / s²

Explanation:

For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use

T = $2\pi \sqrt{ \frac{L}{g} }$

T² = $4\pi ^2 \frac{L}{g}$ 4pi2 L / g

g = $4\pi ^2 \frac{L}{T^2}$

They indicate the average time of 20 measurements 1,823 s, each with an oscillation

let's calculate the magnitude

g = $4\pi ^2 \frac{0.823}{1.823^2}$ 4 pi2 0.823 / 1.823 2

g = 9.7766 m / s²

now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation

for the period

T = t / n

ΔT = $\frac{dT}{dt}$ Δt + $\frac{dT}{dn}$ ΔDn

In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently

ΔT = Δt / n

ΔT = Δt

now let's look for the uncertainty of g

Δg = $\frac{dg}{dL}$ ΔL + $\frac{dg}{dT}$ ΔT

Δg = $4\pi ^2 \frac{1}{T2}$ ΔL + 4π²L (-2 T⁻³) ΔT

a more manageable way is with the relative error

$\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}$

we substitute

Δg = g ( \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}DL / L + ½ Dt / T)

the error in time give us the stanndard deviation

let's calculate

Δg = 9.7766 ( $\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}$ )

Δg = 9.7766 (0.001215 + 0.0184)

Δg = 0.19 m / s²

the absolute uncertainty must be true to a significant figure

Δg = 0.2 m / s2

therefore the correct result is

g ±Δg = (9.8 ± 0.2) m / s²

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3 years ago