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-BARSIC- [3]
3 years ago
7

A 26-kg chandelier hangs from a ceiling on a vertical 4.0-m-long wire. what horizontal force would be necessary to displace its

position 0.17 m to one side?
Physics
1 answer:
-Dominant- [34]3 years ago
8 0
First of all you have that the force that is in cable in its vertical posibion is:
 Fy = m * g
 Fy = (26) * (9.8) = 254.8 N
 Then, the force in horizontal direction will be:
 Tan (x) = Fy / Fx
 Clearing:
 Fx = Fy / tan (x)
 Substituting values:
 Fx = (254.8) / (0.17 / 4)
 Fx = 5995.29 N
 Answer:
 The horizontal force that would be necessary to displace its position 0.17 m to one side is 5995.29 N
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Answer:k = 10.83 N/m²

Explanation: The angular frequency (ω), spring constant (k) and mass is related by the formulae below

ω = √k/m

But ω = 2πf, where f = frequency.

f = number of oscillations /time taken

Number of oscillations = 14, time taken = 11s

f = 14/11 = 1.27Hz.

ω = 2×22/7×1.27

ω = 7.98 rad/s.

By substituting this parameters into ω = √k/m

Where ω = 7.98rad/s, m = 170g = 170/1000 = 0.17kg.

7.98 = √k/0.17

By squaring both sides

(7.98)² = k/ 0.17

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3 years ago
A current of 4 amps is running through a circuit with a resistance of 2ohms.What is the voltage.Show working
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3 years ago
A simple hydraulic lift is made by fitting a piston attached to a handle into a 3.0-cm diameter cylinder. The cylinder is connec
stiv31 [10]

Answer:

Approximately 3.1 \times 10^4 \; \rm N (assuming that the acceleration due to gravity is g = 9.81\; \rm kg \cdot N^{-1}.)

Explanation:

Let A_1 denote the first piston's contact area with the fluid. Let A_2 denote the second piston's contact area with the fluid.

Similarly, let F_1 and F_2 denote the size of the force on the two pistons. Since the person is placing all her weight on the first piston:

F_1 = W = m \cdot g = 50\; \rm kg \times 9.81 \; \rm kg \cdot N^{-1} =495\; \rm N.  

Since both pistons fit into cylinders, the two contact surfaces must be circles. Keep in mind that the area of a square is equal to \pi times its radius, squared:

  • \displaystyle A_1 = \pi \times \left(\frac{1}{2} \times 3.0\right)^2 = 2.25\, \pi\;\rm cm^{2}.
  • \displaystyle A_2 = \pi \times \left(\frac{1}{2} \times 24\right)^2 = 144\, \pi\;\rm cm^{2}.

By Pascal's Law, the pressure on the two pistons should be the same. Pressure is the size of normal force per unit area:

\displaystyle P = \frac{F}{A}.

For the pressures on the two pistons to match:

\displaystyle \frac{F_1}{A_1} = \frac{F_2}{A_2}.

F_1, A_1, and A_2 have all been found. The question is asking for F_2. Rearrange this equation to obtain:

\displaystyle F_2 = \frac{F_1}{A_1} \cdot A_2 = F_1 \cdot \frac{A_2}{A_1}.

Evaluate this expression to obtain the value of F_2, which represents the force on the piston with the larger diameter:

\begin{aligned}F_2 &= F_1 \cdot \frac{A_2}{A_1} \\ &= 495\; \rm N \times \frac{2.25\, \pi\; \rm cm^2}{144\, \pi \; \rm cm^2} \approx 3.1 \times 10^4\; \rm N\end{aligned}.

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3 years ago
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