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3 years ago

Can someone give me some tips of how to study. I have already tried writing everything in my lesson but it doesn't seem to work

1 answer:

3 0

Highlighter, color coding sections that go together, flash cards, relaxing with someone while they read out questions ( this will help with testing anxiety too). If you get distracted easily use games to study or small sayings instead of the usual white and black sheet of paper.

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P*V = n*R*T where P = pressure V = volume n = number of moles R = the universal gas constant T = temperature in degrees Kelvin T

seraphim [82]

The Gay-Lussac's law or Amonton's law states that the pressure of a given amount of a gas is directly propotional to its temperature if its volume is kept constant .

P∝T

and

The Charles Law states that volume of given amount of gas at constant pressure is directly propotional to temperature.

V∝T

So, by Gay-Lussac's law if we increase the temperature the Pressure will increase and by Charles Law, if we increase the temperature the volume will increase.

Therefore, if the temperature of gas increases either the pressure of the gas, the volume of the gas, or both, will increase.

Hence,

Answer is option C

3 0

3 years ago

Read 2 more answers

A ray of light has a wavelength of

MArishka [77]

Answer:

The wavelength in vacuum is equal to 428.8 nm.

Explanation:

Given that,

The wavelength of light, $\lambda=284\ nm$

The refractive index of glass, n = 1.51

We need to find the wavelength in vacuum. The relation between wavelength and refractive index is given by :

$n=\dfrac{\lambda_v}{\lambda}\\\\\lambda_v=n\times \lambda\\\\\lambda_v=1.51\times 284\\\\\lambda_v=428.8\ nm$

So, the wavelength in vacuum is equal to 428.8 nm.

6 0

3 years ago

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the

Zepler [3.9K]

Answer:

51.94°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.19I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}$

The angle between the two filters is 51.94°

5 0

3 years ago

Discuss the relationship between amperage, voltage, and power.

Stells [14]

Electrical power, in watts = (voltage, in volts) x (current, in Amperes)

5 0

3 years ago

A 2.50 kg ball is attached to a 3.00 m bar and swung in a vertical circle. If the ball does not leave the circular loop, what mi

dezoksy [38]

Answer:

5.42 m/s

Explanation:

At minimum speed, the tension in the bar will be 0 when the ball is at the top of the arc, so the only force is gravity pulling down.

Sum of forces towards the center of the circle:

∑F = ma

mg = m v²/r

v = √(gr)

v = √(9.8 m/s² × 3.00 m)

v = 5.42 m/s

6 0

3 years ago