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Tanya [424]
3 years ago
7

What would be the force required to accelerate an object that weighs 35 N up at a rate of .60 m/s^2?

Physics
1 answer:
natta225 [31]3 years ago
3 0

Answer: 2.14\ N

Explanation:

Given

Force required to accelerate the object at the rate of 0.6\ m/s^2

The weight of the object is W=35\ N

Mass of the object can be identified by dividing the weight by acceleration due to gravity

\Rightarrow m=\dfrac{W}{g}\\\\\Rightarrow m=\dfrac{35}{9.8}\\\\\Rightarrow m=3.57\ kg

\therefore \text{Force F=}m\times a\\\Rightarrow F=3.57\times 0.6\\\Rightarrow F=2.14\ N

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A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery
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Answer:

at the beginning: 2.3\cdot 10^{-10} F

when the plates are pulled apart: 1.1\cdot 10^{-10} F

Explanation:

The capacitance of a parallel-plate capacitor is given by

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

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

\epsilon_0 = 8.85\cdot 10^{-12} F/m is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

A=0.210 m^2 is the area of the plates

d=0.815 cm=8.15\cdot 10^{-3} m is the separation between the plates at the beginning

Substituting into the formula, we find

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F

Later, the plates are pulled apart to d=1.63 cm=0.0163 m, so the capacitance becomes

C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F

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