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Aleks04 [339]
4 years ago
6

I NEED THIS ANSWER TODAY PLEASE HELP ME

Physics
2 answers:
Anastaziya [24]4 years ago
7 0

The answer is "B" - If there are no windows then there will be no light coming in, and therefore you don't have to worry about what time of day you do the experiment at.

Inessa05 [86]4 years ago
5 0

B. The time of day


^^ not really sure but haha

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A container with volume 1.83 L is initially evacuated. Then it is filled with 0.246 g of N2. Assume that the pressure of the gas
alisha [4.7K]

Answer:

The  pressure is  P =   1652 \  Pa

Explanation:

From the question we are told that

    The  volume of the container is  V  =  1.83  \ L =  1.83 *10^{-3 } \  m^3

     The mass of  N_2 is  m_n  =  0.246 \ g =  0.246 *10^{-3} \ kg

     The root-mean-square velocity is  v =  192 \ m/s

The  root -mean square velocity is mathematically represented as

      v =  \sqrt{ \frac{3 RT}{M_n  } }

Now the ideal gas law is mathematically represented as

      PV  =  nRT

=>   RT  =  \frac{PV}{n }

Where n is the number of moles which is mathematically represented as

         n =  \frac{ m_n }{M }

Where  M  is the molar mass of  N_2

So  

        RT  =  \frac{PVM_n }{m _n  }

=>    v =  \sqrt{ \frac{3 \frac{P* V  *  M_n }{m_n } }{M_n  } }

=>    v =  \sqrt{  \frac{ 3 *  P* V  }{m_n } } }

=>   P =   \frac{v^2   *  m_n}{3 *    V  }

substituting values

    =>    P =   \frac{( 192)^2   *  0.246 *10^{-3}}{3 *    1.83 *10^{-3} }

=>         P =   1652 \  Pa

       

     

3 0
3 years ago
A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most
alexgriva [62]

Answer:

a) I = 1.44 Kg.m²

b) I = 3.12 Kg.m²

As rarm increases, her moment of inertia will increase.

Explanation:

Given

The skater's mass: M = 64 Kg

Mass of both arms: m = (M/8) = 64 Kg / 2 = 8 Kg

Mass of one arm: = marm = m/2 = 8 Kg / 2 = 4 Kg

Mass of her rest body: Mcyl = M - 2m = 64 Kg - 8 Kg = 56 Kg

rarm = 0.20 m

Rcyl = 0.20 m

Larm = 0.60 m

a) We apply the equation (Using the Steiner Theorem)

I = Icyl + Iarms

I = (Mcyl*Rcyl²/2) + 2*marm*rarm²

I = (56 Kg*(0.20 m)²/2) +2(4 Kg)(0.20 m)²

I = 1.44 Kg.m²

b) Suppose that the center of the mass of her outstretched arm is in the middle, so that the mass of the arm will be at a distance of 50 cm, then

rarm = 0.50 m

We apply the same equation, then

I = Icyl + Iarms

I = (Mcyl*Rcyl²/2) + 2*marm*rarm²

I = (56 Kg*(0.20 m)²/2) +2(4 Kg)(0.50 m)²

I = 3.12 Kg.m²

As rarm increases, her moment of inertia will increase.

8 0
3 years ago
Hy would you expect sodium (Na) to react strongly with chlorine (Cl)?
erastova [34]
Chlorine needs to gain one electron and sodium needs to lose one electron
8 0
3 years ago
An astronaut and his space suit have a combined mass of 157 kg The astronaut is using a soke
alex41 [277]

Answer:

bipartisanship Corp voted for it and it was a deal that firs an u

6 0
3 years ago
Read 2 more answers
Some Physics questions! Please help!
lesantik [10]

1) 260 km/h

Let's use the following convention:

positive x-direction = east

positive y-direction = north

Here we have to find the north component of the velocity's airplane, which means we have to find its y-component.

We can use the formula:

v_y = v sin \theta

where

v = 750 km/h is the magnitude of the plane's velocity

\theta=20.0^{\circ} is the angle between the direction of the plane and the positive x-axis

Substituting,

v_y = (750)(sin 20)=256.5 km/h \sim 260 km/h

2) 24^{\circ} north of east

In order to find the direction of flight, we have to consider that the vector representing the displacement of the plane is the hypothenuse of a right triangle, of which the displacements along the east and north direction are the sides.

Therefore, we have

v_x = 220 km is the displacement towards east

v_y = 100 km is the displacement towards north

Therefore, the angle that gives the direction is given by

tan \theta = \frac{v_y}{v_x}

And substituting,

\theta =tan^{-1}( \frac{100}{200})=24^{\circ}

and this angle is measured north of east.

7 0
4 years ago
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