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Ksju [112]
3 years ago
9

If a lever operates at a mechanical disadvantage, it means that the ________.

Engineering
2 answers:
alekssr [168]3 years ago
4 0

Answer:

The correct answer is option 'B': Load is far from fulcrum and the effort is applied near the fulcrum

Explanation:

A lever works on the principle of balancing of torques. The torque about the fulcrum by the load should be equal to the torque by the applied effort. Since we know that the torque is proportional to both the force and the distance it is applied from the distance from the axis of rotation. A lever is used when we need to lift a heavy load by utilizing this effect of the lever arm.

A mechanical disadvantage occurs when we are not able to lift the weight easily due to the fact we apply effort near the fulcrum.

kherson [118]3 years ago
4 0

Answer:

b.

Explanation:

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Bob and Alice are solving practice problems for CSE 2320. They look at this code: for(i = 1; i <= N; i = (i*2)+17 ) for(k = i
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Answer:

Alice is correct.

The loop are dependent.

Explanation:

for(i = 1; i <= N; i = (i*2)+17 )

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An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

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