They also fashion a gold bullet with a mass of 0.015 kg They fireitfroma gun which has a muzzle velocity of 359 m/s,and the bullet embeds in the wall of their boat:...

A $10.0 \mathrm{g}$ iron bullet with a speed of $4.00 \times 10^{2} \mathrm{m} / \mathrm{s}$ and a temperature of $20.0^{\circ} \mathrm{C}$ is stopped in a $0.500 \mathrm{kg}$ block of wood, also at $20.0^{\circ} \mathrm{C},$ which is fixed in place. (a) At first all of the bullet's kinetic energy goes into the internal energy of the bullet. Calculate the temperature increase of the bullet. (b) After a short time the bullet and the block come to the same temperature $T$. Calculate $T$, assuming no heat is lost to the environment.

Good day. The topic is about heat. We note that when a substance loses or gains heat, its temperature may change. The heat that is associated with temperature change of the substance is solved as Q equals M. C. Delta T. Where Q. Is the heat M. Is the mass C. Is the specific heat which value depends on the type of substance and DELTA T. Is the change in temperature. Which can be solved as the difference between the final temperature T. F. And the initial temperature. The A Let us consider a bullet which mass is three g and moves at the speed of 1 80 m per second. The specific it of the bullet is given to be 1 28 joe's program for a kilogram degree sell shoes. We assume that all of the kinetic energy of the bullet as it moves is transferred to itself in the form of heat. So in this case we wish to find the change in the temperature or by how much the bullet holes temperature will change. So that this first note that since we're using the mass in terms of kilograms to suffer the kinetic energy. So we need to convert 3g into kilograms. And that's done by dividing this by 1000. Since we know that one kg is 1000 g. And that gives us 10000.3 kilogram. So we start by solving for the kinetic energy which is equal to K E kinetic energy one half times mass times the square of the speed. And this gives us 1/2 times the mess .003 times the speed, which is 180 squared. And then Katie will be equal to 48.6. This kinetic energy of Katie will be converted into heat and that it will be observed by the bullet so that the temperature of the bullet will change according to the given equation. So your cue is 48.6. The mass is 0.003. And that specific it is given to be 128, solving for identity gives us 48.6 Over 0.003 times 1 28. And that is equal to yeah, 120 7°C. It follows that the temperature of the bullet will rise up by 127°C. So I hope that helps. Yeah.

In the given question, we have massive plate small chemicals to 0.2 kg and speed of the bullet V equals to 200 meter per second. And we know that the specific heat of the lead bled is 1 34.4 jule per kg. Calvin. Okay, Now, from the energy equation, we can write half when we square. That is, initial energy will be equals to amass identity, identity is converted. Do that bullet has put the question. The complete heat has been absorbed by the bullet only so and will be cancelled from both sides. So Delta T, that is temperature rise will be equals. Two is where they were going to us and substituting the values we get. 200 square. They were by two multiplied by 1 34.4. Okay, so from here, Delta T comes out to be 1 48.8 degrees centigrade. Okay, so this

Hello and welcome to digital tea with mystery, where we will see what's brewing in the world of physics solutions. So in this scenario, we have a lead bullet that shot at a steel plate, which I have shown in the first diagram. And then in the second diagram, we see that the lead bullet hits the plate and embeds itself inside the plate. And one of the assumptions were going to make in this problem is that the heat that is created in that collision between the lead bullet and the steel plate all of that heat is going to remain in the bullet. So then our question is, by how much does the temperature of the bullet increase due to its collision with the steel plate? Uh, excuse me. So, um, let's approach this problem through the conservation of energy. Conservation of energy tells us that the energy in this case before the collision will equal the energy after the collision. So before the collision, the bullet had kinetic energy. And then after the collision, all that Connecticut and he turned into heat. So we have k e turning in the heat so we can write an expression for the K E of 1/2 times MV Squared Mass times velocity squared and that is equal to Ah, the heat, which is mass times specific heat times to change in temperature. And since we're only concerned with the bullet, the masses in in in this case are the same on both sides, so you can cancel them out. So the man said the bullet really doesn't matter. So solving for the change in temperature, I see that I have the V squared over here and it's being multiplied by half, which is the same thing is dividing it by two and then on the other side. I just want to bring over the specific heat so since it's being multiplied on, the right will divide it on the left. So we get the expression of V squared over two times. Two. The specific heat is equal to the change in temperature, so the change in temperature is equal to 275 squared, divided by the quantity of two times 130 when you work that out, you get a grand total of 291 Celsius degrees, so that is How one What least one way you can solve for the change in temperature of an object. It's run through the conservation of energy. Take a look at the type of energy of before and after the collision and in salt for the change in temperature. Hey, thanks for learning with me. I have certainly enjoyed learning with you.