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elena-14-01-66 [18.8K]
2 years ago
13

In the pulley system shown below, a 360 N weight is slowly lifted. Assuming the system is 100% efficient and each pulley is weig

htless, what is the minimum input force needed to lift the weight? Rope Pile L= 16.74m
A. 61 N
B. 121 N
C. 181 N
D. 361 N
Physics
2 answers:
STALIN [3.7K]2 years ago
8 0
Your answer is D. 361 N
Sveta_85 [38]2 years ago
7 0

Answer:

A. 61 N

Explanation:

In this system, there are three moveable pulleys. In this case, the rope will have to move six times as far as the weight will rise. In other words, the input distance is 6 times the output distance  ( d in =6⋅ d out ) . Because the efficiency is 100%, the work input is equal to the work output, so we have:

work input  = work output

f( in)  ⋅  d( in) =   f( out)  ⋅  d( out)

 f( in)  ⋅ (6 ⋅  d( out))  =  f( out)  ⋅  d( out)

 f( in)  ⋅ 6 =  f( out)

  f( in)  ⋅ 6 = 360 N

  f( in)  = 60 N  

Another way to think of this is that since the rope has to move 6 times as far as the height of the weight, the mechanical advantage will be 6. Therefore, the input force needed is  360 ÷ 6 = 60 N . Of course, 60 N will balance the object, so a force slightly greater than 60 N (like 61 N) will be required to lift it.

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An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,
mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

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where would information on the chemical and physical properties of a specific chemical be located in a laboratory or in the work
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Answer:

Both

Explanation:

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Dua titik yang bermuatan listrik yang sama, mula-mula berjarak 5 cm dan saling tarik menarik dengan gaya 225 N. Agar kedua titik
Serggg [28]

Answer:

15 cm

Explanation:

Dari pertanyaan yang diberikan di atas, diperoleh data sebagai berikut:

Gaya 1 (F₁) = 225 N

Jarak terpisah 1 (d) = 5 cm

Gaya 2 (F₂) = 25 N

Jarak terpisah 2 (d₂) =?

Kita dapat memperoleh persamaan yang berkaitan dengan gaya dan jarak muatan dua titik dengan menggunakan rumus berikut:

F = Kq₁q₂ / d²

Perbanyak silang

Fd² = Kq₁q₂

Menjaga Kq₁q₂ konstan, kita memiliki:

F₁d₁² = F₂d₂²

Dengan rumus di atas maka diperoleh jarak sebagai berikut:

Gaya 1 (F₁) = 225 N

Jarak terpisah 1 (d) = 5 cm

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Jarak terpisah 2 (d₂) =?

F₁d₁² = F₂d₂²

225 × 5² = 25 × d₂²

225 × 25 = 25 × d₂²

5625 = 25 × d₂²

Bagilah kedua sisinya dengan 25

d₂² = 5625/25

d₂² = 225

Hitung akar kuadrat dari kedua sisi

d₂ = √225

d₂ = 15 cm

Oleh karena itu, muatan dua titik harus berjarak 15 cm untuk memiliki gaya tarik 25 N

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