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jeyben [28]
3 years ago
15

Can someone answer these?

Physics
1 answer:
scoray [572]3 years ago
5 0

Nine

<em><u>Remark</u></em>

Nine is just a super problem. You understand 9 thoroughly and I guarantee a good mark. It goes to the very heart of physics and the processes it uses.

I don't know if you know about free body diagrams. If you don't, you should. Start with the up and down forces working on the chain pulling upward. Use a circle to represent the center of the action.

Draw an arrow pointing to the top of the page. Label it T for Tension. Now go to the bottom of the circle. Draw two arrows going down. One of them is the boulder. Let the mass of the boulder = M

Let the Mass of the chain = m. Don't worry about the numbers. We'll get around to them later.

So far what you have is T - Mg - mg

Tension is a force.

Mg is a weight which is also a Force.

mg is a weight which is also a Force. So far you have 3 forces. What is the net result? The two weights are accelerated upwards (otherwise the boulder would never be hauled upward. Now we have

<em><u>T - Mg - mg = (m + M)a.</u></em> This is the key equation in the entire problem. We need to use the fact about T not being more that 2.5 times it's own weight.

T ≤ 2.5 * m*g Now we put that into the main equation. We'll use the upper limit. =

2.5mg - (m+M)g = (m + M)a   Now to do just a little math. (No physics). Add (m + M)g to both sides

2.5mg = (m + M)g + (m+M)a  Use the distributive property to factor this

2.5mg = (m + M)(a + g)   Divide by m + M

2.5mg/(m + M) = a + g    We need to isolate a. Subtract g

2.5mg/(m + M) - g = a     We know the entire left side. We can put numbers in at this point. It's not very elegant, but it will work.

m = mass the chain = 575 kg

M = mass of the boulder = 750 kg

g = 9.81 m/s^2

2.5*575*9.81/(575 + 750) - g = a

10.64 - 9.81 = a

a = 0.8329 m/s^2.

Note: The units are correct and the sign on a is correct. The up forces are plus and the down accelerations are minus. Everything is looking good.

The rest of the problem just uses

d = 125 meters

a = 0.8329 m/s^2

vi = 0

t = ???

<em><u>Formula</u></em>

d = vi*t + 1/2 a t^2

125 = 1/2 * 0.8329 t^2  Multiply through by 2 and divide by 0.8329

125 *2 / 0.8329 = t^2

t^2 = 300.1

t = sqrt(300.1)

t = 17.32

Answer: C

Eight

You got this one correct, or we agree that B is the answer. I'll just so a quick solution to this question

This problem is not very well worded. The only way you can get an answer is if you know something about tennis. You have to understand that when you serve, the ball goes straight up. All the forces and velocities are vertical.

Nothing is horizontal until the racket hits the ball. The 73.14 is a horizontal velocity. The question doesn't tell you that and it should

So vi = 0

vf = 73.14

ΔV = 73.14 - 0 = 73.14 m/s

contact time = 30 milliseconds = 30/1000 seconds = 0.03 seconds

mass = W/9.81

<em><u>Formula</u></em>

F = m * ΔV / time

F = (w/9.81) * 73.14  / 0.03

F = w * 248

Answer B

Ten

Ten looks easy, because the math is grade 8 or 9 math. The physics certainly is not. Einstein based some of relativity on a thought problem like this one so the problem is not trivial.

Here's the question. What does the 60N weight represent and which way is it going? Is it going up or down? and if you can decide that, which way is the acceleration (2.45 m/s^2 going, up or down)? Finally who is feeling the 60 N and what does it represent? Now the problem (the physics problem) has presented you with trouble. It is not the math.

The weight of the fish is down.

The 2.45 m/s^2 is up

9.81 of course is down (it always is)

So in order for the elevator to accelerate upward it must be going upward with a net acceleration of 9.81 + 2.45. That's because the elevator must overcome the gravitational acceleration

m*(9.81 + 2.45)  = 60N

m*12.26 = 60N

m = 60/12.26

m = 4.894 kg

Now if the fish were on the ground being held up in the air by a proud fisherman, the fish would actually weigh

F = m*g

m = 4.894

g = 9.81

F = 4.894 * 9.81 = 48 kg

Answer C


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<em>I </em><em>hope </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em><em>have </em><em>a </em><em>nice </em><em>time </em><em>ahead!</em>

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