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Mekhanik [1.2K]
2 years ago
15

A cylinder contains 250 L of hydrogen gas (H2) at 0.0^∘Cand a pressure of 10.0 atm. How much energy is required to raise the tem

perature of this gas to 25.0^∘C?

Physics
1 answer:
Over [174]2 years ago
8 0

Answer:

The amount of energy needed to raise the temperature of the cylinder by 25 °C is 23.3 KJ of heat.

Explanation:

The step by step calculation can be found in the attachment below. Thank you.

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Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks
vivado [14]

The concept of this question can be well understood by listing out the parameters given.

  • The mass of the block = 51 g = 51 × 10⁻³ kg
  • The distance of the block from the table = 30 cm
  • Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

\mathsf{S = ut + \dfrac{1}{2}gt^2}

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

  • Since 1 cm = 0.01 m
  • Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}

\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}

\mathsf{0.3 =4.9*(t^2)}

By dividing both sides by 4.9, we have:

\mathsf{t^2 = \dfrac{0.3}{4.9}}

\mathsf{t^2 = 0.0612}

\mathsf{t = \sqrt{0.0612}}

\mathsf{t =0.247  \ seconds}

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}

where the relation between time (t) and period (T) is:

\mathsf{t = \dfrac{T}{2}}

T = 2t

T = 2(0.247)

T = 0.494 seconds

\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}

By making the spring constant k the subject of the formula:

\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}

\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}

\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}

\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\  \mathsf{  k = \dfrac{2 \pi^2*m}{T^2}}

\mathsf{  k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}

\mathbf{  k =8.25 \ N/m}

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

6 0
2 years ago
A solid cylinder of mass 10 kg is pivoted about a frictionless axis thought the center O. A rope wrapped around the outer radius
taurus [48]
Torque acting dowward = 6  x 0.5 = 3 Nm

Torque acting to the right = 5 x  1 = 5 Nm

5 - 3 = 2 Nm

inertia = 1/2 mr^2

0.5 x 10  x 1^2 = 5 kg-m^2
2/5 = alpha  = 0.4 rad /s^2

Hope this helps
5 0
3 years ago
Read 2 more answers
(b) Show that for certain incident energies there is 100 percent transmission. Suppose that we model an atom as a one-dimensiona
bixtya [17]

Answer:

Explanation:

100 persent transmission implies that the T=1

Therefore using the previous result we have

1+\frac{\sin^2\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a}{4\frac{E}{V_0}\frac{(E+V_0)}{V_0}}=1


\sin\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a=0\Rightarrow \sqrt{\frac{2m}{\hbar^2}(E+V_0)}=0\Rightarrow E=-V_0


The depth of the well for 100% transmission should be

V_0=-0.7~{\rm{eV}}

7 0
2 years ago
Answer the amplitude part of the questions in 7, 8, & 9
erastova [34]
I actually believe for the first question, it would be complete destructive interference as the amplitude and the approximate wavelength for each are the same and will completely or entirely cancel out, rather than simply decreasing or lowering the amplitude as in the bottom question.

The amplitude for the first will be 0, as the 2 waves will cancel each other out. The amplitude of the second, will be 3x, I believe, assuming the amplitude of the first is 2x and the second is 1x, in a constructive interference, I believe the amplitudes would add up.

Likewise for the bottom, I believe you would be subtracting the supposed amplitude of the first which is 2x from 1x which would be 1x.
5 0
3 years ago
One piece of evidence that scientists use to prove that these organisms once existed
VLD [36.1K]
Fossils  Hope that helps!
4 0
3 years ago
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