A friend asks you how much pressure is in your car tires.  Unfortunately, you have misplaced your gauge but confident in your knowledge of physics you tell your friend “I can figure it out”.  From the owner’s manual you find that the car’s mass is 1450 kg and that the tire manufacturer recommends 32 psi.  Assuming that each tire supports one-fourth of the weight of the car, you use a ruler to find that the tires are 15 cm wide and the flattened segment of the tire in contact with the road is 16 cm long.  What answer do you give your friend? Express your answers in pounds per square inch (psi)

OSU has been given a large model of the solar system which consists of hollow Styrofoam spheres of different sizes representing the Sun and planets. The planets are all connected to the center of the sun by means of steel rods of varying lengths and at varying angles relative to the x-axis as shown above. The steel rods are all 0.5 cm in diameter and are made of steel having a density of 7.88 g/cm3. We’d like to suspend the model from it’s center of mass, but it doesn’t look like the center of mass coincides with the position of the Sun. Find the center of mass of the model (direction and angle) relative to the Sun in the center of the model. The Styrofoam spheres are so much lighter in comparison to the rods, that their mass can be neglected. Hint: calculate the center of mass for the x and y dimensions separately, then combine them to get the final answer

Suppose a piano tuner stretches a steel piano wire 7.5 mm. The wire was originally 0.775 mm in diameter, 1.29 m long, and has a Young's modulus of 2.10 × 1011 N/m2.Calculate the force a piano tuner applies to stretch the steel piano wire in Newtons.

Problem 3: A person walks around in high-heeled shoes. Their weight is momentarily placed on one heel which has an area of 1.5 cm². If their mass is 60 kg, what is the pressure exerted by their heel on the floor in pascals?

If the force on the tympanic membrane (eardrum) increases by about 1.5 N above the force from atmospheric pressure, the membrane can be damaged. When you go scuba diving in the ocean, below what depth could damage to your eardrum start to occur. The eardrum is typically 8.2 mm in diameter. In your calculations, use the density of seawater, Psea = 1.03 × 10³ kg/m³.

Pressure cookers have been around for more than 300 years, although their use has strongly declined in recent years (early models had a nasty habit of exploding). How much force (in N) must the latches holding the lid onto a pressure cooker be able to withstand if the circular lid is 30.0 cm in diameter and the gauge pressure inside is 2.60 atm? Neglect the weight of the lid.

A large tub is filled with water, and it is found that the pressure at the bottom is 1) 1.12 x 105 Pa. The same tub is then filled with mercury. What is the new pressure at the bottom of the tub?

A child playing in a swimming pool realizes that it is easy to push a small inflated ball under the surface of the water whereas a large ball requires a lot of force. The child happens to have a styrofoam ball (the shape of the ball will not distort when it is forced under the surface), which he forces under the surface of the water. If the child needs to supply 5.00 × 10² N to totally submerge the ball, calculate the diameter d of the ball. The density of water is pw = 1.000 × 10³ kg/m³, the density of 95.0 kg/m', and the acceleration due to gravity is g = 9.81 m/s². styrofoam is Pfoam =

it’s a lovely day and you decide to launch a giant helium balloon. Once filled, your balloon is a perfect sphere with a radius of 6.90 m. You consider having the balloon lift a crate of water balloons. What is the maximum weight the filled helium balloon can just lift (solve for the equilibrium condition where upward force equals downward force)? Assume that the membrane of the helium balloon has a mass of 920 kg (this does not include the mass of the helium within nor the mass of the crate). Neglect the buoyant force on the crate of water balloons. Density of He is 0.179 kg/m3. Density of air is 1.29 kg/m3. Volume of a sphere is 4/3 π r3. Weight is in N (not kg)

For a normal car riding on tires with relatively flexible sidewalls, the weight of the car is held up, in large measure, by the pressure of the air in the tires. If you look at one of your car’s tires, you’ll note that the tire is flattened slightly to make a rectangle where it touches the ground. The area of the resulting “contact patch” depends on the pressure in the tires. To a good approximation, the upward normal force of the ground (which we can assume is equal to ¼ of the car’s weight) on this patch of the tire is equal to the downward pressure force on the patch.a. Suppose you inflate your 2000 kg car’s tires to the recommended pressure, as measured by a gauge. The resulting contact patch is 18 cm wide and 12 cm long. What does the gauge read?b. If you let a bit of air out of your tire, what happens to the area of the contact patch?