- Pressure = 2000 Pa
- Mass = 100 Kg
- We know, Force = mass × acceleration due to gravity
- Acceleration due to gravity = 9.8 m/s^2
- Therefore, the force = (100 × 9.8) N = 980 N
- We know, Pressure =
- Therefore, area of contact
<u>Answer:</u>
<u>0</u><u>.</u><u>4</u><u>9</u><u> </u><u>m/</u><u>s^</u><u>2</u>
Hope you could understand.
If you have any query, feel free to ask.
The answer to this question is:
Simply it can be calculate using the equation Q=It, here Q is the charge in coulombs , I current in Amps and t is the time in seconds.
so the answer is Q = 10*60 = 600 coulomb.
force in component form is given as
F~ = Fx ˆı + Fy ˆj
given that : Fx = 4 N, Fy = −3 N
hence the force equation looks like
F~ = 4 ˆı + (- 3) ˆj
F~ = 4 ˆı - 3 ˆj
displacement is given as
~s = sx ˆı + sy ˆj
given that : sx = 1 m, and sy = 1 m
hence the displacement is given as
~s = 1 ˆı + 1 ˆj
work done is the dot product of force and displacement . hence work done is given as
W = F~.~s
W = (4 ˆı - 3 ˆj) . (1 ˆı + 1 ˆj )
W = (4 x 1) ( ˆı.ˆı) - (3 x 1) (ˆj.ˆj)
W = 4 - 3
W = 1 J
θ = angle between F~ and ~s = ?
hence the work done by the force comes out to be 1 J
|F~| = magnitude of force = sqrt(Fx² + Fy²) = sqrt((4)² + (-3)²) = 5 N
|~s| = magnitude of displacement = sqrt(sx² + sy²) = sqrt((1)² + (1)²) = sqrt(2)
we know that work done is given as
W = |F~| |~s| Cosθ
1 = (5) (sqrt(2) Cosθ
Cosθ = 0.142
θ = Cos⁻¹(0.142)
θ = 81.84 deg
Answer:
Abdominal closure
Explanation:
The needle has a higher taper ratio and a lower tip angle which makes it sharper. The sharpness (taper-point) of this needle is neccesssary for smooth penetration of tissues such as subcutaneous layers and abdominal viscera, so as to minimize the potential damage of fascia.