Answer:
Explanation:
25 mm diameter
r₁ = 12.5 x 10⁻³ m radius.
cross sectional area = a₁
Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa
a )
velocity of blood v₁ = .6 m /s
Cross sectional area at blockade = 3/4 a₁
Velocity at blockade area = v₂
As liquid is in-compressible
a₁v₁ = a₂v₂
a₁ x .6 m /s = 3/4 a₁ v₂
v₂ = .8m/s
b )
Applying Bernauli's theorem formula
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²
13328 +190.8 = P₂ + 339.2
P₂ = 13179.6 Pa
= 13179 / 13.6 x 10³ x 9.8 m of Hg
P₂ = .09888 m of Hg
98.88 mm of Hg
A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J
Answer:
this vehichle has 896,000 jules of energy
Explanation: KE=1/2mv squared, KE is Kinetic energy. m is mass and v is velocity
Answer:
Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
![W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m](https://tex.z-dn.net/?f=W%3D%5Cint%5Climits%5E%7Bx_f%7D_%7Bx_0%7D%20%7BF%28x%29%7D%20%5C%2C%20dx%20%5C%5CW%3D%5Cint%5Climits%5E%7B14%7D_%7B0%7D%20%28%7B18N-0.530%5Cfrac%7BN%7D%7Bm%7Dx%7D%29%20%5C%2C%20dx%5C%5CW%3D%5B%2818N%29x-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7Bx%5E2%7D%7B2%7D%5D%5E%7B14%7D_%7B0%7D%5C%5CW%3D%2818N%2914m-%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B%2814m%29%5E2%7D%7B2%7D-%2818N%290%2B%280.530%5Cfrac%7BN%7D%7Bm%7D%29%5Cfrac%7B0%5E2%7D%7B2%7D%5C%5CW%3D252N%5Ccdot%20m-52N%5Ccdot%20m%5C%5CW%3D200N%5Ccdot%20m)
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:
The box is initially at rest, so 
. Solving for 
:
