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Dmitriy789 [7]
3 years ago
8

Scientists in a test lab are testing the hardness of a surface before constructing a building. Calculations indicate that the en

tire structure would sink by a certain amount for every additional floor that is added. If the maximum permissible limit for depression of the structure is 20 centimeters, how many floors can be safely added to the building?
Physics
1 answer:
vovikov84 [41]3 years ago
8 0
Let's call this "certain amount" with a variable, x, (since it's not specified in the question, I will assume that the answer shall just be based on this variable).

So every additional floor will make the building sink by x, and let's make the unit cm, so x cm.

Now we know that the maximum depression is 20 cm, and it is also

x*y, where y is the number of floors.

So we have x*y=20,

therefore the number of floors is:
20/x or
\frac{20}{x}



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3 years ago
A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
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Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

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θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

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Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

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