Answer:
Mercury
Explanation:
It is the smallest planet in our solar system and closest to the Sun
m₁ = 2.3 kg <span>
θ₁ = 70° </span><span>
θ₂ = 17° </span><span>
g = 9.8 m/s²
->The component of the gravitational force on m₁ that is parallel down the incline is: </span><span>
F₁ = m₁ × g × sin(θ₁) </span><span>
F₁ = (2.3
kg) × (9.8 m/s²) × sin(70°) = 21.18 N </span><span>
->The component of the gravitational force on m₂ that is parallel down the incline is: </span><span>
F₂ = m₂ × g × sin(θ₂) </span><span>
F₂ = m₂ × (9.8 m/s²) × sin(70°) = m₂ × (2.86 m/s²) </span><span>
Then the total mass of the system is:
m = m₁ + m₂ </span><span>
m = (2.3 kg) + m₂ </span><span>
If it is given that m₂ slides down the incline, then F₂ must be bigger than F₁, </span><span>
and so the net force on the system must be:
F = m₂×(2.86
m/s²) - (21.18 N) </span><span>
Using Newton's second law, we know that
F = m × a
So if we want the acceleration to be 0.64 m/s², then
m₂×(2.86
m/s²) - (21.18 N) = [(2.3 kg) + m₂] ×
(0.64 m/s²) </span><span>
m₂×(2.86
m/s²) - (21.18 N) = (1.47 N) + m₂×(0.64
m/s²) </span><span>
m₂×(2.22
m/s²) = (22.65 N) </span><span>
m₂<span> = 10.2
kg</span></span>
Answer:
True
Explanation:
The tensile stress, σ, on a solid cylindrical wire is given by the following relationship;
Where;
= The tensile force
= The original cross sectional area of the cylindrical wire = π·R²
R = The radius of the wire
Therefore;
= σ × = σ × π × R²
Therefore, the tensile force is directly proportional to the square of the radius of the cylindrical wire, and as the radius of the wire increases, which is by increasing the thickness of the wire, the tensile force is largely increased
The correct option is; True.
