A material's ability to allow heat or electricity to flow through it...

natita [175]

In metric that means multiplying or dividing by 10, 100, 1000, etc. It consists of a partial word like "kilo" or "milli". For example, the prefix "kilo" means "times a thousand" or " one thousand of" the units in question. So "kilometer" means one thousand meters and "kilogram" means one thousand grams.

6 0

4 years ago

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Suppose you find yourself in your friend's third floor apartment building when you smell smoke coming from outside the door. you

Alborosie

(a) Because the tension in the rope would be larger than the maximum value allowed

When you are sliding down the slope, there are two forces acting on you:

- Your weight, $W=mg$ , acting downward, where

m = 70 kg is your mass

g = 9.8 m/s^2 is the acceleration of gravity

- The tension in the rope, T, acting upward

Therefore, the equation of motion is

$T-mg=ma$

where a is the acceleration.

If you want to slide at constant speed, then the acceleration must be zero:

$a = 0$

And so the equation becomes

$T-mg=0$

This means that the tension in the rope must be:

$T=mg=(70 kg)(9.8 m/s^2)=686 N$

Which is larger than the maximum tension allowed in the rope (500 N), so the rope will break.

(b) $2.66 m/s^2$ downward

Again, we must refer to the equation of the forces:

$T-mg=ma$

In this case, we want the tension in the rope to be the maximum allowed value,

T = 500 N

the other data are

m = 70 kg is your mass

g = 9.8 m/s^2 is the acceleration of gravity

Substituting into the equation, we can find the corresponding value of acceleration:

$a=\frac{T-mg}{m}=\frac{500-(70)(9.8)}{70}=-2.66 m/s^2$

where the negative sign means the acceleration is downward.

8 0

4 years ago

I NEED HELP ASAP!!!!!

Annette [7]

Answer:

1. S = d/t S = speed D = distance traveled T= time elapsed

Explanation:

i can answer more in a minute sorry!!

5 0

3 years ago

How much work would it take to push two protons very slowly from a separation of 2.00×10−10 m (a typical atomic distance) to 3.0

Deffense [45]

We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?

The work done is equal to the electric potential energy gained by the proton:

$W=q \Delta V$

where $q=1.6 \cdot 10^{-19}C$ is the charge of the proton and $\Delta V$ is the potential difference between the final position and the initial position of the proton. To calculate this $\Delta V$ , we must calculate the electric potential generated by the proton at rest at the two points, using the formula:

$V=k\frac{Q}{r}$

where $k=9.0 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find