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3 years ago

The volume of a fixed mass of gas is directly proportional to its

Physics

1 answer:

Leni [432]3 years ago

7 0

Answer: Charles's law

Explanation:

Charles's law is one of the gas laws, and it explains the effect of temperature changes on the volume of a given mass of gas at a constant pressure. Usually, the volume of a gas decreases as the temperature decreases and increases as the temperature also increases.

Mathematically, Charles's law can be expressed as:

V ∝ T

V = kT or (V/T) = k

where v is volume, T is temperature in Kelvin, and a k is a constant.

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A car starts from rest at a stop light and reaches 20m/s in 3.5s. Determine the acceleration of the car

olga55 [171]

The car accelerated at around ~5.7 m/s

8 0

3 years ago

State and prove bessel inequality

maria [59]

Statement :- We assume the orthagonal sequence ${{\{\phi\}}_{1}^{\infty}}$ in Hilbert space, now ${\forall \sf \:v\in \mathbb{V}}$ , the Fourier coefficients are given by:

${\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}$

Then Bessel's inequality give us:

${\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Proof :- We assume the following equation is true

${\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}$

So that, ${\bf v_n}$ is projection of ${\bf v}$ onto the surface by the first ${\bf n}$ of the ${\bf \phi_{i}}$ . For any event, ${\sf (v-v_{n})\perp v_{n}}$

Now, by Pythagoras theorem:

${:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}$

${:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}$

Now, we can deduce that from the above equation that;

${:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}$

For ${\sf n\to \infty}$ , we have

${:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Hence, Proved

5 0

2 years ago

Read 2 more answers

Two objects (42.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

Kazeer [188]

Answer:

a = 3.27 m/s²

T = 275 N

Explanation:

Given that:

Mass m₁ = 42.p0 kg

Mass m₂ = 21.0 kg

Consider both masses to be in a whole system, then:

The acceleration can be determined as:

$(m_1+m_2)a = g(m_1-m_2)$

Making acceleration the subject in the above formula;

$a =\dfrac{g(m_1-m_2)}{(m_1+m_2)}$

$a =\dfrac{9.8(42.0-21.0)}{(42.0+21.0)}$

$a =\dfrac{9.8(21.0)}{(63.0)}$

$a =\dfrac{205.8}{(63.0)}$

a = 3.27 m/s²

in the string, the tension is calculated using the formula:

$T = \dfrac{2m_1m_2g}{(m_1+m_2)}$

$T = \dfrac{2(42)(21)(9.81)}{(42+21)}$

$T = \dfrac{17304.84}{63}$

T = 274.68 N

T ≅ 275 N

8 0

2 years ago

Find the magnitude of the sum of two vectors; A is 5 km , and B is 7 , when the angle btween them is 120

Dimas [21]

Answer: 6.24 km

Explanation:

Given

The magnitude of the first vector(say) $\left | a\right |=5\ km$

the magnitude of the second vector(say) $\left | b\right |=7\ km$

the angle between them is $120^{\circ}$

The resultant vector magnitude is given by

$\left | \vec{R}\right |=\sqrt{a^2+b^2+2ab\cos \theta}$

$\left | \vec{R}\right |=\sqrt{5^2+7^2+2\times 5\times 7\cdot \cos 120^{\circ}}\\\left | \vec{R}\right |=\sqrt{74-35}=\sqrt{39}\\\left | \vec{R}\right |=6.24\ km$

4 0

2 years ago

Objects that transmit some light and reflect/absorb the rest of the light, like stained glass, are

11111nata11111 [884]

You are correct it is translucent because opaque let’s no light through but transparent lets almost all light through

3 0

3 years ago