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Ainat [17]
2 years ago
7

How would the photos set up change the mechanical advantage of lifting the block as compared with lifting the block without a le

ver?
A.
B.
C.
D.

Answers in the picture

Physics
2 answers:
Step2247 [10]2 years ago
8 0

I think this will help u

lisabon 2012 [21]2 years ago
5 0
<h3>Answer:</h3>

  The mechanical advantage would decrease, making the block more difficult to lift.

<h3>Explanation:</h3>

The mechanical advantage in such a setup is the ratio of distance from A to B to the distance from D to B. In this picture, that ratio is less than 1, meaning the advantage of having this setup is less than the advantage of no setup at all.

While the force required to lift the block is increased by this setup, the distance over which that force is applied will be smaller for raising the block to a given height. (Overall, for the same height, more work is required with the lever setup because you're raising part of the mass of the lever as well as the mass of the block.)

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An electron is placed on a line connecting two fixed point charges of equal charge but the opposite sign. The distance between t
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Answer:

a)    F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} ),   b) x = 0.15 m

Explanation:

a) In this problem we use that the electric force is a vector, that charges of different signs attract and charges of the same sign repel.

The electric force is given by Coulomb's law

         F =k \frac{q_2q_2}{r^2}

         

Since when we have the two negative charges they repel each other and when we fear one negative and the other positive attract each other, the forces point towards the same side, which is why they must be added.

          F_net= ∑ F = F₁ + F₂

let's locate a reference system in the load that is on the left side, the distances are

left side - electron       r₁ = x

right side -electron     r₂ = d-x

let's call the charge of the electron (q) and the fixed charge that has equal magnitude Q

we substitute

          F_net = k q Q  ( \frac{1}{r_1^2}+ \frac{1}{r_2^2})

          F _net = kqQ  ( \frac{1}{x^2} + \frac{1}{(d-x)^2} )

         

let's substitute the values

          F_net = 9 10⁹  1.6 10⁻¹⁹ 4.50 10⁻⁹ ( \frac{1}{x^2} + \frac{1}{(0.30-x)^2} )

          F_net = 6.48 10⁻¹⁸ ( \frac{1}{x^2} + \frac{1}{(0.300-x)^2} )

now we can substitute the value of x from 0.05 m to 0.25 m, the easiest way to do this is in a spreadsheet, in the table the values ​​of the distance (x) and the net force are given

x (m)        F (N)

0.05        27.0 10-16

0.10          8.10 10-16

0.15          5.76 10-16

0.20         8.10 10-16

0.25        27.0 10-16

b) in the adjoint we can see a graph of the force against the distance, it can be seen that it has the shape of a parabola with a minimum close to x = 0.15 m

4 0
3 years ago
Stan does 178 J of work lifting himself 0.5 m. What is Stan’s mass? The acceleration of gravity is 9.8 m/s 2 . Answer in units o
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Work= (force)(distance)
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178=m(4.905)
178/4.905=m

His mass is 36.3 kg
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2 years ago
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