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fredd [130]
2 years ago
10

The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass

40.0 kg resting on the output plunger. The piston and plunger are nearly at the same height, and each has a negligible mass. By how much is the spring compressed from its unstrained position?
Physics
1 answer:
Triss [41]2 years ago
8 0

Answer:

\Delta x=245\ mm

Explanation:

Given:

  • spring constant of the spring attached to the input piston, k=1600\ N.m^{-1}
  • mass subjected to the output plunger, m=40\ kg

<u>Now, the force due to the mass:</u>

F=m.g

F=40\times 9.8

F=392\ N

<u>Compression in Spring:</u>

\Delta x=\frac{F}{k}

\Delta x=\frac{392}{1600}

\Delta x=0.245\ m

or

\Delta x=245\ mm

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a ball on a string makes 30.0 revolutions in 14.4s, in a circle of radius 0.340m. what is its period.(unit=s)
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A 5.00μF parallel-plate capacitor is connected to a 12.0 V battery. After the capacitor is fully charged, the battery is disconn
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(a) 12.0 V

In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, C=5.00 \mu F, the charged stored on the capacitor at the end of the process is

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When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship

V=\frac{Q}{C}

and since neither Q nor C change, the voltage V remains the same, 12.0 V.

(b) (i) 24.0 V

In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:

C=\frac{\epsilon_0 \epsilon_r A}{d}

where:

\epsilon_0 is the permittivity of free space

\epsilon_r is the relative permittivity of the material inside the capacitor

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d is the separation between the plates

As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: C'=\frac{C}{2}. The new voltage across the plate is given by

V'=\frac{Q}{C'}

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While the separation between the plate was unchanged (d); so, the new capacitance will be

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V'=\frac{Q}{C'}=\frac{Q}{4C}=\frac{1}{4} \frac{Q}{C}=\frac{V}{4}

which means

V'=\frac{12.0 V}{4}=3.0 V

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