<span>R = rate of flow = 0.370 L/s
H = height = 2.9 m
T= time = 3.9 s
V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2
G value = 9.8 m/s2
Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N
Wa = weight of accumulated water after 3.9 s
Fi = force of impact of water on the bucket
S = reading on the scale = Wa + Wb + Fi
mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg
Therefore, Wa = 1.443 x 9.8 = 14.1414 N
Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt)
= 0.37 x 7.539 = 2.78943 N
S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>
When a woman walks south at a speed of 2.0mph for 60 minutes. She then turns around and walks north at a distance of 3000m in 25 minutes. then the woman's average speed during her entire motion would be 73.15 meters /minute.
<h3>What is speed?</h3>
The total distance covered by any object per unit of time is known as speed.
the mathematical expression for speed is given by
speed = total; distance /total time
As given in the problem a woman walks south at a speed of 2.0mph for 60 minutes
60 min = 1 hour
1 mile = 1.60934 km
The distance covered by her southwards = speed ×time
=2 mph × 60 minutes
= 3.218 km
She then turns around and walks north at a distance of 3000m in 25 minutes
The distance covered northward is 3000m
speed = total distance /total time
=(3218 +3000) /(60+25)
=73.15 meters /minutes
Thus, The average speed of the woman would be 73.15 meters /minute.
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Answer:
B. convergent boundary
Explanation:
Convergent boundaries, where ocean plates are subducted at trenches, produce the most tsunamis because they produce the largest earthquakes
Answer:
Explanation:
Give that,
The potential difference of the electrons, 
We need to find the wavelength of the electrons.
Using the conservation of energy,
Put all the values,
So, the wavelength of the electrons is 
.
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
