From the calculation, the gravitational force of attraction is 1.33 * 10^-14 N.
<h3>What is the gravitational force?</h3>
The gravitational force is an attractive force that acts between any two masses.
It is given by;
F = Gm1m2/r^2
F = 6.67 * × 10−11 * 2.5 * 5/(250)^2
F = 83.4 × 10−11 /62500
F= 1.33 * 10^-14 N
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Answer:
Dry lubricants or solid lubricants are materials that, despite being in the solid phase, are able to reduce friction between two surfaces sliding against each other without the need for a liquid oil medium.
Explanation:
first off lemme just say this is really easy man, just look at the directions
Blank #1: -23
Blank #2: 23
A perfectly elastic<span> collision is defined as one in which there is no loss of </span>kinetic energy<span> in the collision. Therefore, we just add the kinetic energies of each system. We calculate as follows:
KE = 0.5(</span>1.0 × 10^3)(12.5 )^2 + 0.5(1.0 × 10^3)(12.5 )^2
KE = 156250 J = 1.6 x 10^5 J -------> OPTION A
Answer:
x = 9.32 cm
Explanation:
For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation
Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar
- W l / 2 - W_{child} x + N₂ l = 0
x = 
1)
now let's use the expression for translational equilibrium
N₁ - W - W_(child) + N₂ = 0
indicate that N₂ = 4 N₁
we substitute
N₁ - W - W_child + 4 N₁ = 0
5 N₁ -W - W_{child} = 0
N₁ = ( W + W_{child}) / 5
we calculate
N₁ = (450 + 250) / 5
N₁ = 140 N
we calculate with equation 1
x = -250 1.50 + 4 140 3) / 140
x = 9.32 cm
