Answer:
Force on the object is 20 N
Explanation:
As we know that work done to raise the book from initial position to final position is known as potential energy stored in it
So here we know that
here we know that
U = 30 J
s = displacement = 1.5 m
so we have
Answer:
The answer is ""
Explanation:
Please find the complete question in the attached file.
pi = pressure only at two liquids' devices
PA = pressure atmosphere.
1 = oil density
2 = uncertain fluid density
The pressures would be proportional to the quantity cm from below the surface at the interface between both the oil and the liquid.
Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 210kPa + 101.325kPa
P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)
T = 25°C = 298.15K
Final P and T values:
P = ?, T = 0°C = 273.15K
Set the initial and final P/T values equal to each other and solve for the final P:
311.325/298.15 = P/273.15
P = 285.220kPa
Subtract 101.325kPa to find the final gauge pressure:
285.220kPa - 101.325kPa = 183.895271kPa
The final gauge pressure is 184kPa or 26.7psi.
Answer: d boiling water
Explanation: the water is boiling and bubbling cause the molecules are so fast and they hit off each other
Answer:
The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)
Explanation:
Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.
Step 2: I must calculate the magnitude of the forces acting on the third charge.
F13: Force exerted by charge 1 on charge 3.
F23: Force exerted by charge 2 on charge 3.
K: Constant of Coulomb's law.
d13: distance from charge 1 to charge 3.
d23: distance from charge 2 to charge 3
Fr: Resulting force.
q1=+2.06 x 10-9 C
q2= -3.27 x 10-9 C
q3= +1.05 x 10-9 C
K=9-10^9 N-m^2/C^2
d13= 0,20 m
d23= 0,10 m
F13= K * (q1 * q3)/(d13)^2
F13=9,7335*10^(-8) N
F23=K * (q2 * q3)/(d23)^2
F23= -3,09 * 10^(-7)
Step 3: We calculate the resultant force on charge 3.
Fr=F13+F23= -2,11665 * 10^(-7)