Answer:
[tex]P = 2.94 \times 10^5 Pa[/tex]
Explanation:
Normal force due to four tires is counter balancing the weight of the car
So here we will have
[tex]4F_n = mg[/tex]
[tex]F_n = \frac{mg}{4}[/tex]
[tex]F_n = \frac{1.20 \times 10^3 \times 9.81}{4}[/tex]
[tex]F_n = 2943 N[/tex]
now we know that pressure in each tire is given by
[tex]P = \frac{F}{A}[/tex]
Here we know that
[tex]A = 1.00 \times 10^2 cm^2 = 1.00 \times 10^{-2} m^2[/tex]
[tex]P = \frac{2943}{1.00 \times 10^{-2}}[/tex]
[tex]P = 2.94 \times 10^5 Pa[/tex]
Answer:
P = 294300Pa or 42.67psi by conversion.
Explanation:
Since Four tyres were inflated, we have that area of the four tyres are
4×1×10²cm²
Pressure is given as:
P = f/a but f = mg
P = m×g/a
Therefore,
P = 1.20x10³kg × 9.81m/s² / (4 ×1x10² cm²)
P = 1.20x10³kg×9.81m/s² / (0.04m²)
P = 294300Pa or 42.67psi by conversion.
