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Mars2501 [29]
3 years ago
11

The load is the pivot point of a lever.

Engineering
2 answers:
iren2701 [21]3 years ago
7 0

Answer:

False

Explanation :

lever is a rigid bar which turn around a pivot point, called a fulcrum. It is a simple machine and can be use to move a load (or do work). Effort is applied to one part of the lever and it pushes the load at another end. Therefore, the load is not the lever part of the fulcrum or pivot.

olganol [36]3 years ago
6 0

Answer:Its False!

Explanation:

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Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's activities.

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What happens to a commercial airline at cruising altitude if the pilot does not touch the throttles?
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3 years ago
Analyze the example of this band saw wheel and axle. The diameter of the wheel is 14 inches. The diameter of the axle that drive
Kazeer [188]

The answer for the ideal mechanical advantage and actual mechanical advantage for the different scenarios are;

A) Ideal Mechanical Advantage = 18.67

B) Actual Mechanical Advantage = 4.1067

We are given;

Input distance; The diameter of the wheel; d_w = 14 inches

Output distance; The diameter of the axle that drives the wheel; d_a = 3/4 inches

The force needed to cut a one-inch-thick softwood board; F = 1.75 pounds

The efficiency of the band saw; η = 22% = 0.22

A) Formula for Mechanical advantage is;

M.A = Force output/Force input = (Input distance)/(Output distance)

Thus;

Ideal mechanical advantage = 14/(3/4)

Ideal mechanical advantage = 18.67

B) Now, we are given that efficiency of the band saw is η = 22% = 0.22.

Thus using the mechanical advantage formula above;

Actual mechanical advantage = 0.22 × Expected output

Actual mechanical advantage = 0.22 × 18.67

Actual mechanical advantage ≈ 4.1067

Read more about Mechanical Advantage at; brainly.com/question/18345299

5 0
2 years ago
A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2
Svetllana [295]

79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

Shrinkage limit is lowest water content

e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

7 0
3 years ago
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