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4 years ago

Which are causes of mechanical weathering? (check all that apply)

Physics

2 answers:

solmaris [256]4 years ago

8 0

The correct answers are:

B. plant growth;

C. animal actions;

The mechanical weathering is a type of weathering where physical force is included into the breaking up of the rocks. The plants and the animals are both causing this type of weathering with their actions. The plants can cause mechanical weathering with their roots, as they grow and surround a rock, they are able to create such a pressure that they can break the rock apart. Also, as their trunks are getting bigger, if there's rocks right next to them, the pressure from the growing of the trunk will crack the rocks. The animals are able to move the rocks, as well as pushing them, or even deliberately throwing them, so they manage to break up parts of them and cause mechanical weathering.

Hatshy [7]4 years ago

6 0

The answers are Plant growth, animal actions, and pressure release.

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Tlo

kondor19780726 [428]

Answer:

9.949 m

Explanation:

From the question,

L' = L+LαΔT................. Equation 1

Where L' = New length, L = Original length, α = linear expansivity of Cu, ΔT = change in temperature

Given: L = 10 m, α = 0.00017 K⁻¹, ΔT = 50-80

L' = 10+10(0.00017)(50-80)

L' = 10-0.051

L' = 9.949 m

Hence the new legth of Cu is 9.949 m

5 0

3 years ago

Which of the following material offers the highest resistivity?

klasskru [66]

The answer would be gold

8 0

3 years ago

An object weighs 100 newtons on earth.What is its weight on the moon?

Pachacha [2.7K]

Answer:

a) The gravitational acceleration at the surface of the Moon is g moon=1.67 m/s

The ratio of weights (for a given mass ) is the ratio of g-values, so

moon

=(100N)(1.67/9.8)=17N.

(b) For the force on that object caused by Earth's gravity to equal 17 N, then the free fall acceleration at its location must be

=1.67m/s

. Thus , .

= r 2

⇒

=1.5×10

So the object would need to be a distance of r/R

=2.4 "radii" from Earth's center.

4 0

3 years ago

Read 2 more answers

A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill.

zloy xaker [14]

Answer:

$P=28.085\,hp$

Explanation:

Given that:

mass of 1 skier, $m=80kg$

inclination of hill, $\theta=25^{\circ}$

length of inclined slope, $l=220m$

time taken to reach the top of hill, $t=2.3 min= 138 s$

coefficient of friction, $\mu=0.15$

Now, force normal to the inclined plane:

$F_N=m.g.cos\theta$

$F_N=80\times 9.8\times cos25^{\circ}$

$F_N=710.54\,N$

Frictional force:

$f=\mu.F_N$

$f=0.15\times 710.54$

$f=106.58\,N$

The component of weight along the inclined plane:

$W_l=m.g.sin\theta$

$W_l=80\times 9.8\times sin25^{\circ}$

$W_l=331.33\,N$

Now the total force required along the inclination to move at the top of hill:

$F=f+W_l$

$F=106.58+331.33$

$F=437.91\,N$

Hence the work done:

$W=F.l$

$W=437.91\times 220$

$W=96340.80\,J$

Now power:

$P=\frac{W}{t}$

$P=\frac{96340.80}{138}$

$P=698.12\,W$

So, power required for 30 such bodies:

$P=30\times 698.12$

$P=20943.65\,W$

$P=\frac{20943.65}{745.7}$

$P=28.085\,hp$

8 0

3 years ago

Starting from rest, a small block of mass m slides frictionlessly down a circular wedge of mass M and radius R which is placed o

guapka [62]

Answer:

Part a)

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

$mv + MV = 0$

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR$

now we have

$\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR$

$\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR$

so we have

$V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

and another block has speed

$v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}$

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

$y_{cm} = \frac{mR}{m + M}$

7 0

3 years ago