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agasfer [191]
3 years ago
6

A 10 Newton effort force is required to lift a 10 kg box just off the ground. What is the effort required to lift the box off th

e ground using a 3m long lever? [The fulcrum is placed 0.5m from the box.]
Physics
2 answers:
Zanzabum3 years ago
6 0

Answer:

effort required to  lift the box off the ground using a 3m long lever=2N

Explanation:

In this question we have given

force required,F=10N

mass of box.m=10kg

length of lever, l=3m

distance between fulcrum and box,d=.5m

effort required to  lift the box off the ground using a 3m long lever=?

Here,

Effort required to  lift the box off the ground using a 3m long lever \times (3m-0.5m) =F \times 0.5m

Effort required to  lift the box off the ground using a 3m long lever \times (3m-0.5m) =10N \times 0.5m

Effort=\frac{10\times 0.5}{3-0.5}

Effort=2N

Karo-lina-s [1.5K]3 years ago
5 0

2.0 N is the answer, I hope this helps :D




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Answer:

If all these three charges are positive with a magnitude of 1.8 \times 10^{-8}\; \rm C each, the electric potential at the midpoint of segment \rm AB would be approximately 8.3 \times 10^{3}\; \rm V.

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\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}.

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