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3 years ago

7

A lever can lift 65 N easily. How would you change the lever so that it can lift 150 N? You would increase mechanical advantage

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1 answer:

scoray [572]3 years ago

4 0

<span>You would increase mechanical advantage by moving the fulcrum closer to the output source (load).</span>

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On a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.

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Answer:

The correct answer is - 8.99N/C

Explanation:

$dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k$

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.

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3 years ago

An engineer measured the length of a building which was equal to 55.2 m. If there was an error of 0.02 m, what are the probabili

Vinvika [58]

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https://web.lemoyne.edu/courseinformation/mth%20112/rinaman/instman/SOLUTION.PDF

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2 years ago

how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at

DiKsa [7]

Answer:

$|a|=2.83\ m/s^2$

$\theta=75^o$

Explanation:

<u>Net Force And Acceleration </u>

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

$\vec F_n=m\vec a$

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

$\vec F_1=$

$\vec F_1=\ N$

The components of the second force are

$\vec F_2=$

$\vec F_2=\ N$

The net force is

$\vec F_n=$

$\vec F_n=\ N$

The magnitude of the net force is

$|F_n|=\sqrt{14.64^2+54.64^2}$

$|F_n|=\sqrt{3200}=56.57\ N$

The acceleration has a magnitude of

$\displaystyle |a|=\frac{|F_n|}{m}$

$\displaystyle |a|=\frac{56.57}{20}$

$|a|=2.83\ m/s^2$

The direction of the acceleration is the same as the net force:

$\displaystyle tan\theta=\frac{54.64}{14.64}$

$\theta=75^o$

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2 years ago