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frozen [14]
3 years ago
12

Chinese public buildings erected under a construction code of the Sung dynasty have withstood earthquakes well because the white

cedar used has four times the tensile strength of steel and the timber frame, incorporating many joints and few nails, is flexible:_________.(A) used has four times the tensile strength of steel and the timber frame, incorporating.(B) used in them has four times the tensile strength of steel has and the timber frame, incorporating.(C) that was used in them has four times the tensile strength steel has, and the timber frame, incorporating.(D) that was used has four times as much tensile strength as steel, and the timber frame incorporates.(E) that was used has four times the tensile strength steel does, and the timber frame incorporates.
Physics
1 answer:
luda_lava [24]3 years ago
8 0

Answer:

A: used has four times the tensile strength of steel and the timber frame, incorporating

Explanation:

Option A is correct because it conveys the correct message intended by the statement and has no grammatical errors.

Option B is wrong because to say "has four times the tensile strength of steel has" is just grammatically and idiomatically wrong as has is used twice in the sentence.  

Option C is wrong because the statement that has to do with the flexibility of the timber's frame is more like a separate fact and does not fall under the scope of trying to further explain a fact.

Option D is wrong because it has the same problem in Option C. The comma that is placed after "steel" breaks the sentence and hence does not provide a good understanding of why the building can withstand earthquakes.

Option E is grammatically wrong for using does in the sentence "has four times the tensile strength steel does"

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Write the function of force​
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2 years ago
Two charged metallic spheres of radii, R = 10 cms and R2 = 20 cms are touching each other. If the charge on each sphere is +100
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Answer:

200\times 10^{-6}j

Explanation:

We have given the radius of first sphere is 10 cm and radius of second sphere is 20 cm

So the potential of first sphere will be greater than the potential of the second sphere, so charge will flow from first sphere to second sphere

Let q charge is flow from first sphere to second sphere and then potential become same

So V=\frac{K(100-q)}{r_1}=\frac{K\times 100}{r_2}

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So V=\frac{K(100-q)}{r_1}=\frac{9\times 10^{9}\times (100-33.33)\times 10^{-9}}{10\times 10^{-2}}=6003V

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8 0
2 years ago
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Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?
irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

1)

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

M=Fd

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

M_c = W_2 d_2

where W_2 = 500 N is the weight of the person and d_2 = 2 m is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

M_a = W_1 d_1

where W_1 is his weight and d_1 = 4 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the person on the left:

W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N

2)

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

M_c = W_2 d_2

where W_2 = 400 N is the weight of the person and d_2 is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 300 N is his weight and d_1 = 4 m is the distance from the fulcrum.

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the distance of the person on the right:

W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m

3)

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

M_c = W_2 d_2

where W_2 is the weight of the seesaw and d_2 = 3 m is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 600 N is his weight and d_1 = 1 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the seesaw:

W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N

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