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Alexxx [7]
3 years ago
15

An ideal gas is at a pressure 1.00 Ã 105 n/m2 and occupies a volume 2.00 m3. if the gas is compressed to a volume 1.00 m3 while

the temperature remains constant, what will be the new pressure in the gas?
Physics
1 answer:
blsea [12.9K]3 years ago
4 0
The behavior of an ideal gas at constant temperature obeys Boyle's Law of
p*V = constant
where
p = pressure
V = volume.

Given:
State 1:  
  p₁ = 10⁵ N/m² (Pa)
  V₁ = 2 m³
State 2:
  V₂ = 1 m³

Therefore the pressure at state 2 is given by
p₂V₂ = p₁V₁
or
p₂ = (V₁/V₂) p₁
    = 2 x 10⁵ Pa

Answer: 2 x 10⁵ N/m² or 2 atm.
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Sallys physical education teacher timed her run and recorded the time and distance in the table below. What is her average speed
Fudgin [204]
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1 + 2 + 3 + 4 + 5 = 15
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36 divided by 5 = 7.2


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4 0
3 years ago
Please help me to solve this and give me the summary of answer
Neko [114]
Using the principle of floatation.

u = w............(a)

Upthrust of fluid is equal to the weight of the object.

Let the volume of the wood be V.

The upthrust u, is related to the volume submerged in water, and that is 1/5 of it volume, that is (1/5)V = 0.2V

Formula for upthrust, u = vdg

where v = volume of fluid displaced
d = density of fluid
g = acceleration due to gravity

weight, w = mg
where m = mass
g = acceleration due to gravity

From (a)

                     u = w

                 vdg =  mg      Cancel out g

                   vd  =  m
 
The v  is equal to 0.2V, which is the submerged volume. Notice that the small letter v is volume of fluid displaced, and capital V is the volume of the solid.

d is density of fluid which is water in this case, 1000 kg/m³

         0.2V * 1000 =  m

           200V =  m

Hence the mass of the object is  200V  kg.

But Density of solid =  Mass of solid / Volume of solid

                                 =    200V / V

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7 0
3 years ago
Enumerate the viscosity of magma in different conditions.
Firlakuza [10]

Answer:

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3 0
2 years ago
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º. What is the coefficient of
Alex_Xolod [135]

Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

To Find :

The coefficient of static friction between the box and the plane.

Solution :

Vertical component of force :

mg\ sin\ \theta =  120\times 10 \times sin\ 47^\circ{}=877.62 \ N

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mg\ cos\ \theta =  120\times 10 \times cos\ 47^\circ{}=818.40 \ N

Since, box is on the verge of slipping :

mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07

Therefore, the coefficient of static friction between the box and the plane is 1.07.

Hence, this is the required solution.

7 0
3 years ago
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