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Romashka-Z-Leto [24]
3 years ago
8

If the effort distance = 8 meters and the resistance distance = 1/2 meter. How much force does this man have to apply to lift th

e 1 ton (2000 N) weight?
Physics
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

Explanation:

Effort x effort distance = load x resistance  distance

effort distance = 8 m ,

load = 2000N

resistance distance = 1/2 m = 0.5 m

Putting the values in the equation above

effort x 8m = 2000N x .5

effort = 2000 x 0.5 / 8

= 125 N

force required = 125 N .

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Answer:

  • Current = 0.33 A

Explanation:

  • For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

\dashrightarrow \: \:  \sf V = 2 \times 5 = 10 V

Three resistor of 5\Omega, 10\Omega, 15\Omega are connected in Series, so the net resistance:

\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}

\dashrightarrow \: \:  \sf R = 5 + 10 + 15

{ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \:  \Omega}}}}

According to ohm's law:

\dashrightarrow  \sf\: \: V = IR

\dashrightarrow  \sf \: \: I = \dfrac{V}{R}

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 {\pmb{\sf{\Omega}}} we get:

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{ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}

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4 0
2 years ago
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bixtya [17]

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but

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So we have

u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}

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6 0
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