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Bezzdna [24]
2 years ago
13

A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa

lly by the two tires. If the pressure in each tire is 7.60 x 10^5 Pa, what is the area of contact between each tire and the ground?
Physics
1 answer:
kozerog [31]2 years ago
7 0

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

given,                                  

weight of the person = 625 N

weight of the bike = 98 N        

Pressure on each Tyre = 7.60 x 10⁵ Pa

Area of contact on each Tyre = ?          

total weight of the system = 625 + 98

                                             = 723 N

Let F be the force on both the Tyre

F + F = W                                    

2 F  = 723                                    

F = 361.5 N                                

F = P A                                            

A = \dfrac{F}{P}                          

A = \dfrac{361.5}{7.60 \times 10^5}

A = 4.76 x 10⁻⁴ m²

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Answer:

height of the water rise in tank is 10ft

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substitute P_a_t_m for P_1, (P_a_t_m +pgh) for P_2

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\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}

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\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4

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\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}

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Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

\sqrt{64.4(20-h)} for V₂

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A₂V₂ = A₃V₃

A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft

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