An aluminium cube of side 5cm and R. D 2.7 is suspended by a thread in alcohol of relative density 0.8o. Find the tension in thread.​

The tension in the thread is approximately 3.2103225 Newtons

Explanation:

The given parameters are;

The side length of the aluminum cube, s = 5 cm

The R.D. of aluminum = 2.7

The relative density of alcohol into which the aluminum cube is placed = 0.8

The volume of the aluminum cube, V = s³ = (5 cm)³ = 125 cm³

The mass of the aluminum cube, V = The volume of the cube

The mass of the aluminum cube, m = V × R.D. × Density of water

m₁ = 0.000125 × 2.7 × 970 = 0.327375 kg

The mass of the alcohol displaced, m₂ = 0.000125 × 0.8 × 970 = 0.097

The tension in the string, T = The apparent weight of the aluminum cube, Wₐ

∴ The tension in the thread, T = Wₐ = (m₁ - m₂) × g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ T ≈ (0.327375 - 0.000125) × 9.81 = 3.2103225

The tension in the thread, T ≈ 3.2103225 N.

d) the number of neutrons plus the number of protons

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards