Answer:
a)906.5 Nm^2/C
b) 0
c) 742.56132 N•m^2/C
Explanation:
a) The plane is parallel to the yz-plane.
We know that
flux ∅= EAcosθ
3.7×1000×0.350×0.700=906.5 N•m^2/C
(b) The plane is parallel to the xy-plane.
here theta = 90 degree
therefore,
0 N•m^2/C
(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
therefore, applying the flux formula we get
3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C
Answer:
3675 J
Explanation:
Gravitational Potential Energy = 
× mass × g × height
( g is the gravitation field strength )
Mass = 50 kg
G = 9.8 N/kg ( this is always the same )
Height = 15 m
Gravitational Potential Energy = × 50 ×9.8 × 15
= 3675 J
Answer:
400m
Explanation:
Brainliest? :))
Let your initial displacement from your home to the store be
Dd
>
1 and your displacement from the store to your friend’s house
be Dd
>
2.
Given: Dd
>
1 = 200 m [N]; Dd
>
2 = 600 m [S]
Required: Dd
>
T
Analysis: Dd
>
T 5 Dd
>
1 1 Dd
>
2
Solution: Figure 6 shows the given vectors, with the tip of Dd
>
1
joined to the tail of Dd
>
2. The resultant vector Dd
>
T is drawn in red,
from the tail of Dd
>
1 to the tip of Dd
>
2. The direction of Dd
>
T is [S].
Dd
>
T measures 4 cm in length in Figure 6, so using the scale of
1 cm : 100 m, the actual magnitude of Dd
>
T is 400 m.
Statement: Relative to your starting point at your home, your
total displacement is 400 m [S].
Answer:
26.9 Pa
Explanation:
We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:
(1)
where
is the cross-sectional area of the 1st section of the pipe
is the cross-sectional area of the 2nd section of the pipe
is the velocity of the 1st section of the pipe
is the velocity of the 2nd section of the pipe
In this problem we have:
is the velocity of blood in the 1st section
The diameter of the 2nd section is 74% of that of the 1st section, so
The cross-sectional area is proportional to the square of the diameter, so:
And solving eq.(1) for v2, we find the final velocity:
Now we can use Bernoulli's equation to find the pressure drop:
where
is the blood density
are the initial and final pressure
So the pressure drop is:
