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soldier1979 [14.2K]
3 years ago
14

HELP PLS WHAT IS Y COMPONENT

Physics
1 answer:
jeka943 years ago
6 0

Answer:

11.5

Explanation:

you do 12(sin 73.3) because its on the y you do sin

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A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr
jasenka [17]

Answer: The original temperature was

T_{1}=126.51K

Explanation:

Let's put the information in mathematical form:

V_{1}=12.7m^{3}

T_{1}=?

V_{2}=32.5m^{3}

T_{2}=323K

P_{1}=P_{2}=3atm

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

PV=nRT

were <em>R</em> is the gas constant. And <em>n</em> is the number of moles (which we don't know yet)

From this, taking R=0.08205746\frac{atm.l}{mol.K},  we have:

n=\frac{P_{2}V_{2}}{RT_{2}}

⇒n=3.67mol

Now:

T_{1}=\frac{P_{1}V_{1}}{nR}

⇒T_{1}=126.51K

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2 years ago
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What if is a forecasting game, below are actions on your document and all you have to do is predict what will happen after the a
Talja [164]

1. Triple click on the paragraph.

By extension, the triple click can allow you to select a line and the quadruple click a section (paragraph).

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If you press one of the arrow keys with the Shift key pressed, you can select a section of text.

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2 years ago
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Can you guys help to understand this graph I'm so confused why I getting wrong?
dedylja [7]
The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.

Segment D ...
Walking AWAY from home; distance increases as time increases.

Segment B ...
Not walking; distance doesn't change as time increases.

Segment C ...
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2 years ago
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A 10.00 kg mass is moving to the right with a velocity of 14.0 m/s. A 12.0 kg mass is moving to the left with a velocity of 8.00
Basile [38]

Answer:

2 m/s

Explanation:

From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.

Let the 10.00 kg mass be m_1 and the 12.0 kg mass be m_2. When they collide and stick, they have a combined mass of m_1+m_2.

Momentum is given by p=mv. Set up the following equation:

\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_f, where v_f is the desired final velocity of the masses.

Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.

Solving yields:

10\cdot 14 + 12\cdot (-8)=(10+12)v_f\\\implies v_f=\boxed{2 \text{ m/s}}

4 0
2 years ago
Find deacceleration An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When
earnstyle [38]

Answer:

The deceleration is  a = -0.7273 \ m/s^2  

Explanation:

From the question we are told that

  The distance of the car from the crossing is d = 360 \  m

   The speed is u  = 16 \  m/s

    The reaction time of the engineer is  t = 0.53 \ s

Generally the distance covered during the reaction time is  

      d_r = u * t

=>   d_r = 16 * 0.53

=>   d_r = 8.48 \ m

Generally distance of the car from the crossing after the  engineer reacts is

      D = d- d_r        

=> D = 360 - 8.48      

=> D = 352 \ m

Generally from kinematic equation

      v^2 = u^2 + 2as

Here v is the final velocity of the car which is  0 m/s

So

        0^2 = 16^2 + 2 * a * 352

=>    a = -0.7273 \ m/s^2  

3 0
2 years ago
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