Answer:
pressure of the water = 3.3 × 
pa
Explanation:
given data
velocity v1 = 1.5 m/s
pressure P = 400,000 Pa
inside radius r1 = 1.00 cm
pipe radius r2 = 0.5 cm
h1 = 0 (datum at inlet)
h2 = 5.0 m (datum at inlet)
density of water ρ = 1000 kg/m³
to find out
pressure of the water
solution
we consider here flow speed in bathroom that is = v2 and Pressure in bathroom is = P2
here we will use both continuity and Bernoulli equations
because here we have more than one unknown so that
v1 × A1 = v2 × A2 × P1 + ρ g h1 + (0.5)ρ v1² = P2 + ρ g h2 + (0.5) ρ v2²
now we use here first continuity equation for get v2
v2 = 
v2 = 
v2 = 6 m/s
and now we use here bernoulli eqution for find here p2 that is
P2 = P1 - 0.5× ρ ×(v2² - v1²) - ρ g (h2- h1 )
P2 = 400000 - 0.5× 1000 ×(6² - 1.5²) - 1000 × 9.81 × (5-0 )
P2 = 3.3 × pa
Answer:
Distance covered is: 45 meters
Displacement is 15 meters to the right of where he started
Explanation:
Notice that Brady has walk a path that looks like an incomplete rectangle of height 5 meters and length 25meters, although he actually didn't cover the full length (25 meters) when getting back to the point where he started (he made just 10 meters instead of 25 after the third turn right) See attached image.
Therefore, Brady's displacement is 15 meters to the right of where he started, and the total distance he covered is :
Distance = 5m + 25m + 5m + 10m = 45m
Answer:
D.
Explanation:
I think its D. because Tea during a cold workout isnt safe at all. Drinking tea while working out adds weight and you just simply dont need tea when working out.. Hope this helps man :)
Answer:
5.65 times
Explanation:
60 db sound is equal to 60 phons sound when frequency is kept at 1000Hz.
But when the frequency of sound is changed to 100 Hz , according to equal loudness curves , the loudness level on phon scale will be 35 phons.
A decrease of 10 phon on phon- scale makes sound 2 times less loud
Therefore a decrease of 25 phons will make loudness less intense by a factor equal to 2²°⁵ or 5.65 less intense . Therefore intensity at 100 Hz
must be increased 5.65 times so that its intensity matches intensity of 60 dB sound at 1000 Hz frequency.
From my research, the image supports the question. From the graph given, we can construct the equation of the line using the two-point formula. Using the given value of 601 K, we can solve for the missing value of altitude.
y - y1 = [(y2 - y1)/(x2 - x1)](x- x1)
y - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4)
Substituting y = 601 to solve for x:
601 - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4<span>)
</span>x = 83
Therefore, the probe's instruments will fail at 83 kilometers.
