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4 years ago

Consider three scenarios in which a particular box moves downward under the pull of gravity :

Physics

1 answer:

luda_lava [24]4 years ago

5 0

Answer:

1. True WA > WB > WC

Explanation:

In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.

A) Work is the product of force by distance and the cosine of the angle between them

WA = W h cos 0

WA = mg h

B) On a ramp without rubbing

Sin30 = h / L

L = h / sin 30

WB = F d cos θ

WB = F L cos 30

WB = mf (h / sin30) cos 30

WB = mg h ctan 30

C) Ramp with rubbing

W sin 30 - fr = ma

N- Wcos30 = 0

W sin 30 - μ W cos 30 = ma

F = W (sin30 - μ cos30)

WC = mg (sin30 - μ cos30) h / sin30

Wc = mg (1 - μ ctan30) h

When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum

Let's review the claims

1. True The work of gravity is the greatest and the work where there is friction is the least

2 False. The job where there is friction is the least

3 False work with rubbing is the least

4 False work with rubbing is the least

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Can rock undergo compression, tension, and shear stress all at once? explain

Katena32 [7]

This causes reverse faults, which are the reverse of normal faults, because in this case, the hanging wall slides upward relative to the footwall. Shear stress is when rock slabs slide past each other horizontally. There is no vertical movement of either the hanging wall or footwall, and we get a strike-slip fault.

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3 years ago

A 25 kg circular disk has a diameter of 2.5 feet and a thickness of 2.5 cm. Find the density of the disk in kg/m3. Next, find th

Gre4nikov [31]

Answer:

Assume that $\rm g= 9.81\; N\cdot kg^{-1}$ ; $\rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}$ .

Density of the disk: approximately $\rm 2.19\times 10^{3}\; kg\cdot m^{-3}$ .

Weight of the disk: approximately $\rm 245\;N$ .

Buoyant force on the disk if it is submerged under water: approximately $\rm 112\; N$ .

The disk will sink when placed in water.

Explanation:

Convert the dimensions of this disk to SI units:

Diameter: $d = \rm 25\; inches = (25\times 0.3048)\; m = 0.762\;m$ .
Thickness $h = \rm 2.5\; cm = (2.5\times 0.01)\; m = 0.025\;m$ .

The radius of a circle is 1/2 its diameter:

$\displaystyle r = \rm \frac{1}{2}\times 0.762\;m = 0.381\; m$ .

Volume of this disk:

$V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}$ .

Density of this disk:

$\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}$ .

$\rho(\text{disk}) >\rho(\text{water})$ indicates that the disk will sink when placed in water.

Weight of the object:

$W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N$ .

The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately $\rm 0.0114009\; m^{3}$ of water. The buoyant force on the disk will be:

$\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}$ .

The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.

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3 years ago

Which of think following materials is the best is the best choice for an insulating material?

RSB [31]

You are correct, the answer is C.

Good insulators are materials that do not allow heat, electricity, light or sound pass through easily. Materials like steel and silver (which are metals) are not good insulators because they make good conductors. They easily allow heat, electricity and the like pass through them.

Thus, the answer is letter C. Rubber.

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3 years ago

What is the process when beach sediment moves down the beach with the current?

Solnce55 [7]

The answer is:
________________________________________
"longshore drift" .
___________________________________________

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4 years ago

A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a

Fynjy0 [20]

Answer:

d= 7.32 mm

Explanation: