Answer:a) P = Po + rho×h×g
b) P = 5.4 × 10^9 pa
c) F = P/A = (Po + rho×h×g)/A
d) 1.174×10^11N
Explanation: Using the formula
P = Po + rho×h×g
P = 1.0 x 10^5 + 1000 × 5.5 × 9.81
P = 5.4 × 10^9pa
The magnitude of the force exerted by water on the top of the person's head F at the depth h in terms of P
F = P/A = (Po + rho×h×g)/A
Using the above formula
Where A = 0.046m^2
F = P/ A = 5.4×10^9/0.046
F = 1.174×10^11N
Answer:
None, if air resistance is ignored.
Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
Vertical acceleration due to gravity affects the vertical velocity by accelerating the object toward the center of the earth, and by decreasing the upward vertical velocity..
The horizontal component of velocity makes the object travel horizontally as long as the projectile is airborne.
Thsi discussion assumes that air resistance is ignored.
Answer:
Spiders cannot actually propel their bodies through the water as a swimmer does, but they can use objects to get across the water and some can run across the water.
Explanation:
Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
