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Aleksandr [31]
2 years ago
7

A planet has a surface temperature of 95.0 K and an atmospheric pressure of 1.60 atm. If 4.87 L of atmosphere has a mass of 28.6

g, what is the average molar mass of the atmosphere? Use R=0.08206L atmmol K for the gas constant. Your answer should include three significant figures.
Physics
1 answer:
mr Goodwill [35]2 years ago
3 0

Answer:

M=28.88 gm/mol

Explanation:

Given that

T= 95 K

P= 1.6 atm

V= 4.87 L

m = 28.6 g

R=0.08206L atm .mol .K

We know that gas equation for ideal gas

P V = n R T

P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant

Now by putting the values

P V = n R T

1.6 x 4.87  = n x 0.08206 x 95

n=0.99 moles

We know that number of moles given as

n=\dfrac{m}{M}

M=Molar mass

n=\dfrac{m}{M}

0.99=\dfrac{28.6}{M}

M=28.88 gm/mol

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IgorC [24]

Answer:

2.98 second

Explanation:

The severity index is defined by :

S=a^{5/2}t

a is dimensionless constant that equals the number of multiples of g

Conditions are given as :

Initial velocity, u = 0

Acceleration, a = 34 m/s²

Final velocity, v = 16.4 km/h = 4.56 m/s

We can find t from the above data as follows :

t=\dfrac{v-u}{a}\\\\t=\dfrac{4.56-0}{34}\\\\t=0.134\ s

As a is the acceleration that is multiple of g.

So,

a=\dfrac{34}{9.8}=3.46

So,

Severity index,

S=a^{5/2}t\\\\S=(3.46)^{5/2}\times 0.134\\\\S=2.98\ s

Hence, the severity index for the collision is 2.98 seconds.

6 0
2 years ago
Three balls with the same radius 21 cm are in water. Ball 1 floats, with
Masteriza [31]

Answer:

Explanation:

A )

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B )

Tension in the ball will be equal to net force acting on the ball

Net force on the ball = buoyant force - weight .

4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )

= 40.65 x 10⁻⁶ N .

C )Tension in the 3 rd  ball will be equal to net force acting on the ball

Net force on the ball =  weight  - buoyant force

= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000  )

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7 0
2 years ago
The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha
katovenus [111]

Answer:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

  • When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
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5 0
2 years ago
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Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot
Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

                  m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0

                  (m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0

Taking out ‘mg’ as common and we get

                  2 x-50=0

                  2 x=50

                  x=\frac{50}{2}=25 \mathrm{cm}

Hence, the stick should be pivoted at a distance of,

                 x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

                   4 m g(x)+(m g)(x-50 c m)=0

                   4 m g x+m g x-50 m g=0

Taking out ‘mg’ as common and we get

                   5 x=50

                   x=\frac{50}{5}=10 \mathrm{cm}

Hence, the stick should be pivoted at a distance of,

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So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

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2 years ago
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Likurg_2 [28]

Answer

Given,

Periscope uses 45-45-90 prisms with total internal reflection adjacent to 45°.

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refractive index of glass, n_g = 1.52

When the light enters the water, water will act as a lens and when we see the object from the periscope the object shown is farther than the usual distance.

7 0
2 years ago
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