Question

The question is attached to this. i would like a full answer and explanation ASAP.
thank you!​

Answer 1

(A) The pressure in both tire is 17.97 N/cm^2 or 179,667 Pascals or 26.1 psi.

(B) It is difficult to predict what will happen with the area of the tires in contact with the Earth the event of a puncture. Higher, perhaps, since the tire flattens out, adding surface area.

Step-by-step explanation:

Given

55kg

15 cm^2 per tire, with 2 tires

g = 9.8 N/kg

Calculate the force of the bicycle on the Earth

Force = Mass*Acceleration

Force = (55kg)*(9.8 N/kg)

Force = 539 N

Calculate the force per cm^2 contact area

Area = 2*(15 cm^2) = 30 cm^2

Pressure = 539N/30 cm^2

Pressure = 17.97 N/cm^2

Change to a standard SI unit for pressure, the Pascal [For fun, the problem doesn't specify units]

Since 1 Pascal = 1 N/m^2, and

1 cm^2 = 0.0001 m^2

Pressure = (17.97 N/cm^2)*(1 cm^2/0.0001 m^2)

Pressure = 179,667 Pascals

For the non-SI crew

1 Pascal = 0.000145 psi

Pressure = 26.1 psi

(A) The pressure in both tire is 17.97 N/cm^2 or 179,667 Pascals or 26.1 psi.

(B) It is difficult to predict what will happen with the area of the tires in the event of a puncture. The flattened tire will spread out a little, so one might say the contact area increases. But the intent of the question is not clear. Good luck.

Answer 2

(a) 1796.67 N/m²

(b) If there's a puncture in one of the tires, the tire's contact area with the ground will increase as the tire flattens due to the decreased internal pressure. This change helps distribute the weight over a larger area but can negatively impact the bicycle's performance

Step-by-step explanation:

We are asked to answer the following:

(a) Calculate the pressure in the tires.

(b) Explain what happens to the area of the tires if one of the tires starts losing pressure.

To address these parts, we'll first calculate the pressure needed in the bicycle tires to support the bicycle and the rider, and then explore the effects of a puncture on the contact area of the tire with the ground.

We are given:

• A = 15 cm²
• g = 9.8 N/kg

Now let's solve.

`\hrulefill`

Answering Part (a):

Pressure is defined as force per unit area. Since the force is distributed over the area of both tires, the total area involved is twice the area of one tire.

`\boxed{\left\begin{array}{ccc}\text{\underline{Pressure:}}\\\\ P = (\vec F)/(A)\\ \\\text{Where:}\\\text{$\bullet$ $P$ is the pressure in $(N)/(m^2)$ } \\ \text{$\bullet$ $\vec F$ is the force exerted}\\\text{$\bullet$ $A$ is the area over which the force is exerted}\end{array}\right}`

The total weight of the bicycle and rider is the product of the total mass and the gravitational field strength. This weight is the force exerted on the ground. Let's calculate this:

`\Longrightarrow \vec w=mg\\\\\\\\\Longrightarrow \vec w=(55 \ kg)(9.8 \ (N)/(kg))\\\\\\\\\therefore \vec w = 539 \ N`

Thus, the bike and rider weigh 539 N.

Now we can use our formula for pressure. Remember, the total area involved is twice the area of one tire. We also have to convert from cm² to m². Let's now calculate the pressure:

`\Longrightarrow P = (539 \ N)/(2(0.15 \ m^2))\\\\\\\\\therefore P=\boxed{1796.67 \ (N)/(m^2)}`

Thus, the calculated pressure in the bicycle tires needed to support the bicycle and the rider is approximately 1796.67 N/m².

Answering Part (b):

When a tire is punctured, the pressure inside it begins to decrease. Pressure is what keeps the tire rigid and maintains its shape, so a decrease in pressure will cause the tire to deform more under the weight of the bicycle and rider. This deformation increases the contact area of the tire with the ground. This phenomenon helps distribute the weight over a larger area, although it adversely affects the bicycle's performance and efficiency.