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worty [1.4K]
3 years ago
13

A block of mass 3 kg slides down a frictionless inclined plane of length 6 m and height 4 m. If the block is released from rest

at the top of the incline, what is its speed at the bottom?
Physics
1 answer:
Ludmilka [50]3 years ago
6 0
(if it's frictionless the length doesn't even matter :) )
It would have the same kinetic energy down as the potential energy up. That is, mgh=\frac{mv^2}{2} or 2gh=v^2 (the mass doesn't even matter). The result is \sqrt{2gh}, so only the height matters really. It is almost 9 (it is \sqrt{80}=4\sqrt{5}).
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4 0
2 years ago
How can i find the acceleration?(rope and grinder have no weight) *** sorry for my english
Finger [1]
The main formula to be used here is

                       Force = (mass) x (acceleration).

We'll get to work in just a second.  But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them.  So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.

On the <u>right</u> side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration.  I don't know what to do with that, because . . .

At the <em>same time</em>, both of those will be pulled <u>up</u> by the 10 kg on the other side
of the upper pulley.

I think I can handle the 10 kg, and work out the acceleration that IT has.

Let's look at only the forces on the 10 kg:

-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.

-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope. 

So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.

The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.

The acceleration of 10 kg, with 49 newtons of force on it, is

     Acceleration = (force) / (mass) = 49/10 = <em>4.9 meters per second²</em>
7 0
3 years ago
Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 1.4×109 electrons from one disk to the
allsm [11]

Answer:

r = 6.5*10^-3 m

Explanation:

I'm assuming you meant to ask the diameters of the disk, if so, here's it

Given

Quantity of charge on electron, Q = 1.4*10^9

Electric field strength, e = 1.9*10^5

q = Q * 1.6*10^-19

q = 2.24*10^-10

E = q/ε(0)A, making A the subject of formula, we have

A = q / [E * ε(0)], where

ε(0) = 8.85*10^-12

A = 2.24*10^-10 / (1.9*10^5 * 8.85*10^-12)

A = 2.24*10^-10 / 1.6815*10^-6

A = 1.33*10^-4 m²

Remember A = πr²

1.33*10^-4 = 3.142 * r²

r² = 1.33*10^-4 / 3.142

r² = 4.23*10^-5

r = 6.5*10^-3 m

3 0
3 years ago
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