Answer: 0.031rad/s
Explanation:
Tangential speed = 50.0m/s
Diameter = 3200m
Radius = diameter/2
= 3200/2 = 1600m
Angular speed = tangential speed / radius
= 50/1600
= 0.031rad/s
Answer:
Therefore the velocity of the proton is m/s towards east.
Explanation:
The mass of proton is m= kg.
q= C is the magnitude of the charge of a proton.
The earth's magnetic field is B= T toward north.
Let v be the velocity of the proton towards east.
The angle between the velocity of proton towards east and magnetic field towards north is .
We know that,
Now putting the all values
m/s
Therefore the velocity of the proton is m/s towards east.
The correct answer is C.
We will use Boyle's law that states that for a fixed amount of an ideal gas kept at a fixed temperature, pressure and volume are inversely proportional.
P1 V1 = P2 V2
Where
P1 is initial pressure = 5 psi
V1 is initial volume = 20 cubic inch
P2 is final pressure = 10 psi
V2 is final volume = unknown
V2 = P1,V1 / P2
V2 = 20 × 5 / 10
V2 = 100/10
V2 = 10 cubic inches
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr