Answer:
Work is done in moving a charge of 2 coulomb across two points having a potential difference of 12 volt is 24 joule .
Explanation:
Answer:
w= 62.75 J
Explanation:
Given that
Force vector F= 5 x i + 4 y j
Space vector or displacement vector d= 5.01 i
We know that work (w)
w=∫ F.ds
w= ∫(5 x i + 4 y j) .dx ( only object is moving in x- direction)
![w=\left [\dfrac{5}{2}x^2\right ]_0^{5.01}](https://tex.z-dn.net/?f=w%3D%5Cleft%20%5B%5Cdfrac%7B5%7D%7B2%7Dx%5E2%5Cright%20%5D_0%5E%7B5.01%7D)
w= 2.5 x 5.01² J
w= 62.75 J
If John goes running around his neighborhood everyday after work, the health-related factors that he will be improving through this type of training is any factors that involve the improvement of cardiorespiratory systems. This would improve a lot such as his circulation, endurance, stamina, and etc. Hope this helps.
Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa
