Explain how you would draw magnetic field lines in the vicinity of this magnet (it is a normal north and south magnet)

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$

This is the new external F force that we are looking for:

$F=Friction1(ground)+Friction2(AB)$

we know Friction2(AB) because we found that in the previous block so:

$F=Friction1(ground)+mg$ *μs

for the other friction we have to use the equation:

$Friction(ground)=N(ground)$ *μs

from y axis we have:

$N(ground)-w-Normal(AB)=0$

$N(ground)=w+Normal(AB)$

we found the value of Normal(AB) with the previous block so:

$N(ground)=mg+mg=2mg$

and:

$Friction(ground)=2mg$ *μs

$F=Friction(ground)+mg$ *μs

$F=2mg$ *μs+μs* $mg=3mg$ *μs

and since: μs* $mg=41.6N$

the new F force would be:

$F=3mg$ *μs $=41.6*3=124.8N$

4 0

3 years ago

What is an example of a mechanical noise?

faltersainse [42]

Resonance:
The resounding recurrence is the recurrence at which a bit of metal, plastic or whatever else swings/vibrates with minimal measure of vitality input. Think about a man on a play area swing. You realize that it requires next to no push to keep the individual swinging. The recurrence at which they swing forward and backward is their full recurrence. In the event that you endeavor to influence them to swing speedier or slower, it will take altogether more vitality.

Resonating Panels:
This kind of clamor is caused when the bass notes are an indistinguishable recurrence from the thunderous recurrence of a metal or plastic board. To stop or decrease the commotion related with this kind of issue, you can do two or three things.

Rattling:
This sort of commotion would be caused when 2 bits of metal, plastic, whatever... are sufficiently close to hammer into each other when they resound. This is most likely best illuminated by filling the hole between the two vibrating parts with silicone sealant or shut cell froth climate stripping. The climate stripping is a superior arrangement in places like behind the tag. On the off chance that you have a tag outline, you can get some truly thin climate stripping and put between the casing and the plate.

4 0

3 years ago

Which situation would deplete freshwater?

sasho [114]

Answer:

If freshwater consumption was greater than freshwater renewal.

Explanation: