<em>So</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>4</em><em>k</em><em>g</em><em>/</em><em>m^</em><em>3</em><em>.</em>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em>
<em>G</em><em>ood</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
Answer:
Minimum uncertainty in velocity of a proton,
Explanation:
It is given that,
A proton is confined to a space 1 fm wide,
We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,
Since, p = mv
So, the minimum uncertainty in its velocity is greater than . Hence, this is the required solution.
<h3>
Answer: Approximately 4.67 m/s^2</h3>
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Explanation:
Let's say you want to push the brick to the right. The free body diagram will have an arrow pointing right on the rectangle (the brick) and the arrow is labeled with 35 N.
Friction always counteracts whatever force you apply. The friction force arrow will point left and be labeled with 7 N.
The net horizontal force is therefore 35-7 = 28 N and the direction is to the right. The positive net force means you've overcome the force of friction and the brick is moving.
F = 28 is the net force
m = 6 is the mass
a = unknown acceleration
F = m*a .... newton's second law
28 = 6a
6a = 28
a = 28/6
a = 4.67
The acceleration of the brick is approximately 4.67 m/s^2
This means that for every second, the brick's velocity is increasing by about 4.67 m/s.
Answer:
The lowest possible frequency of sound for which this is possible is 1307.69 Hz
Explanation:
From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.
First, we will determine his distance from the second speaker using the Pythagorean theorem
l₂ = √(2.00²+5.00²)
l₂ = √4+25
l₂ = √29
l₂ = 5.39 m
Hence, the path difference is
ΔL = l₂ - l₁
ΔL = 5.39 m - 5.00 m
ΔL = 0.39 m
From the formula for destructive interference
ΔL = (n+1/2)λ
where n is any integer and λ is the wavelength
n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.
Then,
0.39 = (1+ 1/2)λ
0.39 = (3/2)λ
0.39 = 1.5λ
∴ λ = 0.39/1.5
λ = 0.26 m
From
v = fλ
f = v/λ
f = 340 / 0.26
f = 1307.69 Hz
Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.