A 15.0 kg cart is moving with a velocity of 7.50 m/s down a level hallway. A constant force of 10.0 N acts on the cart, and its
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Answer:
The pressure difference will increase by the factor of 1.75
Explanation:
For constant flow rate, coefficient of viscosity, length of the vessel and the pressure difference is inversely proportional to the fourth power of the radius of the blood vessel
Apply the principle of Poiseuille’s law.
Q = (P2 - P1)/R
Pls check the attached file for step by step solution of the question. It is submitted in this way as typing the equation may not be explanatory.
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
