Mg = 6.2 x 9.81 = 60.822
This is also normal force.
Coefficient of friction times normal force is the force due to friction:
60.822 x 0.24 = 14.6N
F = MA so F(your force) - F(friction) = 6.2 x 0.5
= 3.1
Your answer is 3.1+ 14.6
I hope this is correct though I might be wrong.
Answer:
3.83×10¯⁴ N
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +2.4x10¯⁸ C
Charge 2 (q₂) = +1.8x10¯⁶ C
Distance apart (r) = 1.008 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
The magnitude of the electrical force acting between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × 2.4x10¯⁸ × 1.8x10¯⁶ / (1.008)²
F = 0.0003888 / 1.016064
F = 3.83×10¯⁴ N
Thus the magnitude of the electrical force acting between the two charges is 3.83×10¯⁴ N
Explanation:
its the minimum amount of energy required to remove the most loosely bound electron
To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.
By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore




Here,
m = mass
g =Gravitational energy
The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

Remember the expression for which you can determine the relationship between mass, volume and density, in which

In this case the density would be that of the object, replacing

Since the displaced volume of water is 0.429 we will have to


The density of water under normal conditions is
, so


The density of the object is 