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
1 answer:
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
M=Molar mass
M=28.88 gm/mol
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Answer:
A. Attractive
B. ( μ₀I² ) / ( 2πd )
Explanation:
A. We know that currents in the same direction attract, and currents in the opposite direction repel, according to ampere's law. In this case the current in the two wires are flowing in the same direction, and hence the force between the two wires are attractive.
B. Suppose that two wires of length and
both carry the current
in the same direction ( given ). In the presence of a magnetic field produced by wire 1, a force of magnitude m say, is experienced by wire 2. The magnitude of the magnetic field produced by wire 1 at distance say d, from it's axis, should thus be the following -
= μ₀I / 2πd
The force experienced by wire 2 should thus be -
= I(
)
= I
Sin( 90 )
= I
( μ₀I / 2πd )
Therefore the force per unit length experienced by wire 2 toward wire 1 should be ...
( /
) = ( μ₀I² ) / ( 2πd ) ... which is our solution
The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>
All moving particles with mass have a matter wave associated with it. These matter waves are called deBroglie waves.
The deBroglie wavelength λ of a particle is given by,
Here, h is the Planck's constant, m is the mass of the ball and v is its velocity.
Calculate the deBroglie wavelength of the moving golf ball by substituting 6.626×10⁻³⁴J s for h, 45.9×10⁻³kg for m and 62.0 m/s for v.
The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>