Answer:
Explanation:
D = 8.27 m ⇒ R = D / 2 = 8.27 m / 2 = 4.135 m
ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s
We can apply the equation
Ff = W ⇒ μ*N = m*g <em>(I)</em>
then we have
N = Fc = m*ac = m*(ω²*R)
Returning to the equation <em>I</em>
<em />
μ*N = m*g ⇒ μ*m*ω²*R = m*g ⇒ μ = g / (ω²*R)
Finally
μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379
Density is defined as [mass] / [volume] .
The only choice listed with those physical dimensions is 'd' .
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass
kg, charge +e =
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed
m/s. The proton comes momentarily to rest at a distance
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are
m apart?
Explanation:
The given data is as follows.
Mass of proton =
kg
Charge of proton = 
Speed of proton = 
Distance traveled = 
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
=

where, 
U = 
Putting the given values into the above formula as follows.
U = 
= 
= 
Therefore, we can conclude that the electric potential energy of the proton and nucleus is
.
Answer:
The letter B is the letter that represents the location of the resister in the diagram.
Explanation:
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Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m