7200 grams is what equivalen to 7.s kilograms
Answer:
38.47 m
Explanation:
To find the height of the building, we will use the following equation
Where yf is the final height, yi is the initial height, viy is the initial vertical velocity, t is the time, and a is the acceleration due to gravity.
If the brick is in flight for 3.1 s, we can say that when t = 3.1s, yf = 0 m. So, replacing
viy = (16 m/s)sin(10) = 2.78 m/s
a = -9.8 m/s²
we get
Solving for yi
Therefore, the height of the building is 38.48 m
Answer:
C) gravity separated from the unified force, strong force separated from the unified force, inflationary expansion occurred, electromagnetic and weak forces separated from the unified force, quarks and electrons formed
Answer:
1.686 m
Explanation:
From coulomb's law,
F = kq1q2/r² ...................................... Equation 1
Where F = electrostatic force between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.
making r the subject of the equation,
r = √(kq1q2/F).......................... Equation 2
Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C
Constant: k = 9.0×10⁹ Nm²/C².
Substituting into equation 2
r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)
r = √(14364×10⁻³/5.05)
r = √(14.364/5.05)
r = √2.844
r = 1.686 m
r = 1.686 m.
Thus the distance must be 1.686 m
