Y8 III TRI Summary PDF

Y8 III TRI summary Jedhai Pimentel  —— Y8 III TRI summary —— F uids In science, a fluid is any substance that can flow. Both liquids and gases are classif ed as fluids because their parti...

Y8 III TRI summary Jedhai Pimentel  —— Y8 III TRI summary —— F uids In science, a fluid is any substance that can flow. Both liquids and gases are classif ed as fluids because their particles can move freely past one another. For example, water (a liquid) and air (a gas) are both fluids, even though they behave differently. Temperature Temperature is a measure of how agitated the molecules are. The faster the particles move, the higher the temperature. Temperature determines how hot or cold something is and can be measured in different units, such as Celsius (°C), Fahrenheit (°F), or Kelvin (K). Temperature influences the state of matter (solid, liquid, gas) and affects the behavior of particles, such as their speed and energy. It also affects density, viscosity, volatility, etc. Viscosity Viscosity is the internal friction of a fluid. It refers, in practical terms, to how thick or thin a liquid is and how easily it flows. A liquid with high viscosity, like honey, flows slowly because its particles are more resistant to motion. A liquid with low viscosity, like water, flows easily. Viscosity decreases with increasing temperature because the particles move faster and can slip past each other more easily. The opposite is also true, when temperature decreases viscosity increases. 2 l i —— Y8 III TRI summary —— Volume Volume is the amount of space that an object or substance occupies. It can refer to solids, liquids, or gases, but in gases, volume is related to the container’s size, since gases expand to f ll the space available. Pressure Pressure is the force exerted per unit area on the surface of an F object, so we use the formula P =. In gases, pressure results from A collisions of gas particles with the walls of their container. Pressure increases when the force on a surface increases or when the area over which the force is distributed decreases. Density Density is how well the mass of something is spread among its volume. m Thats why the formula for it is mass divided by volume (ρ = ), and V it determines whether an object will float or sink in a fluid. If an object’s density is less than the fluid’s, it will float, but if it’s greater, it will sink. 3 i —— Y8 III TRI summary —— Buoyancy Buoyancy is the upward force exerted by a fluid, which allows objects to float. The buoyant force is equal to the weight of the fluid displaced by the object. It is calculated as B = ρ ⋅ g ⋅ VD , where ρ is the density of the fluid, g is gravity (always 10 on Earth) and VD is the volume of fluid displaced by the object (that means the volume of the object that is underwater). Weight Weight is the downward force that everything feels, because we are in Earths gravitational f eld. When an object floats in a fluid is because the weight (downwards) and the buoyancy (upwards) cancel each other out. We calculate weight as w = m ⋅ g, where m is the mass of the object and g is gravity (always 10 on Earth). Temperature, Pressure and Volume If the temperature of a gas remains constant, increasing the volume of the gas decreases its pressure and vice versa. This is because pressure is inversely proportional to volume. If the volume doubles, the pressure is halved. If temperature is not constant, this law is Pi ⋅ Vi Pf ⋅ Vf expressed as = , where P is pressure, V is volume, and Ti Tf T is temperature (the little i stands for “initial" and the little f P⋅V for “f nal". This means that, for a sealed container remains T constant. 4 i i —— Y8 III TRI summary —— Thermal Expansion and contraction When a gas is heated inside a sealed container, the particles move faster and collide with the walls more frequently, increasing the pressure. This can cause the container to expand or even rupture if the pressure becomes too high. This is why heating a gas in a sealed container leads to an increase in pressure. The opposite is also true, so when we decrease the temperature, the pressure will also decrease. 5 —— Y8 III TRI summary —— Exercises 1. You leave a sealed bottle of soda in a hot car. After a few hours, the bottle looks like it’s about to burst. What’s happening inside the bottle in terms of gas pressure and temperature? 2. You’re helping your family clean the pool, and you notice that a plastic toy floats while a metal key sinks. Why do these objects behave differently in water? 3. You’re watching a science experiment where a syringe with a gas inside has the plunger pulled back to increase the volume. What will happen to the pressure it the temperature doesn’t change? Why does this happen? 4. Imagine you’re designing a device to transport gases at high altitudes where the temperature is very low. How does temperature affect the pressure inside the gas containers during the trip? 5. You’re in a swimming pool with a large inflatable ball. When you try to push the ball underwater, it keeps floating back up. What forces are acting on the ball, and why does it resist submersion? 6. You f nd a cube with a volume of 1 and a mass of 2. Calculate its density. If you drop the cube in a swimming pool, will it float or sink? Explain your reasoning. 7. You’re inflating a balloon and the volume increases from 2 to 4. If the temperature remains constant, and the initial pressure was 200, what is the new pressure inside the balloon? 8. You have a container of gas with a volume of 10 and a pressure of 150 at room temperature (25). If the temperature is increased to 100 while keeping the volume constant, what is the new pressure inside the container? 9. You have a container of gas with a volume of 10 and a pressure of 300 at room temperature (30). If the temperature is increased to 90 while the volume increases to 15, what is the new pressure inside the container? 10.While snorkeling, you f nd a stone with a a volume of 2. Calculate the buoyant force acting on the stone when it’s fully submerged in water. 6 i i —— Y8 III TRI summary —— 11.You are building a small miniature boat using a box that has a base area of 2. If the mass of the plus its cargo is 50, what is the minimum height of the box needed to keep it afloat? 12.You have a ball with a radius of 0.2 m and a mass of 2 kg. The ball floats on water, and you need to calculate the minimum force required to push it completely underwater. Use π = 3, g = 10 and the volume of a sphere 4 can be calculated as Vs = ⋅ π ⋅ r 3. 3 7 —— Y8 III TRI summary —— Solutions 1. In the hot car, the gas inside the bottle expands as it heats up, increasing the pressure and making the bottle swell. 2. The plastic toy floats because it is less dense than water, while the metal key sinks because it is denser than water. 3. Since temperature is constant, volume and pressure are inversely proportional. That way, as the volume increases the pressure will decrease. That happens because pressure depends on the area, and when we increase the volume there is more area available for the molecules of the gas to colide with the walls of the container. 4. At high altitudes, where the temperature is low, the gas particles move slower. That way, they collide less with the walls of the container, which means that pressure decreases. 5. The buoyant force is pushing the ball upward, while weight is pulling it down. Since the ball’s density is lower than water’s, it floats. 2 6. The density is ρ = = 2. Since the density is greater than that of water 1 (1), the cube will sink. 7. Because the temperature is constant: P1 ⋅ V1 = P2 ⋅ V2 200 ⋅ 2 = 4 ⋅ P2 400 P2 = 4 P2 = 100 8 —— Y8 III TRI summary —— 8. Because the volume is constant: P1 P = 2 T1 T2 150 P = 2 25 100 P2 = 100 ⋅ 6 P2 = 600 9. Now that nothing is constant, we need to use the full formula: P1 ⋅ V1 P ⋅V = 2 2 T1 T2 300 ⋅ 10 P ⋅ 15 = 2 30 90 P2 300 = 6 P2 = 6 ⋅ 300 = 1800 10.Using the formula (note the VD = 2 because it is fully submerged) B = ρ ⋅ g ⋅ VD B = 1 ⋅ 10 ⋅ 2 B = 20 9 : —— Y8 III TRI summary —— 11.The buoyant force must equal the weight of the boat, so: B=w ρ⋅g⋅V =m⋅g ρ⋅V =m m V= ρ 50 V= = 50 1 We know that the volume is (Ab is the base area and h is the height) V = Ab ⋅ h V h= Ab 50 h= = 25 2 12.To calculate this, we need to know what is the buoyancy of the ball fully submerged, because this is the value that we need to push down to put the ball underwater, so: B =ρ⋅g⋅V Since we don’t know the volume of the ball, we must calculate it: 4 Vs = ⋅ π ⋅ r3 3 4 Vs = ⋅ 3 ⋅ 23 3 10 : —— Y8 III TRI summary —— Vs = 4 ⋅ 8 = 32 Now we go back to the buoyancy: B =ρ⋅g⋅V B = 1 ⋅ 10 ⋅ 32 = 320 We don’t need to make the full force, because weight is helping us push it down, so: w =m⋅g w = 20 ⋅ 10 = 200 So the force that we need to make is: F = B−w F = 320 − 200 = 120 11