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Sophie [7]
3 years ago
14

Which electrical protective device is designed to detect a difference in current between circuit wires and interrupt the circuit

and stop the flow of electricity? (1 point)
a. GFCI
b. Circuit breaker
c. Fuse
d. Insulation
Physics
1 answer:
Tamiku [17]3 years ago
6 0
The electrical protective device is designed to detect a difference in current between circuit wires and interrupt the circuit and stop the flow of electricity is : A.GFCI
It stands for Grand Fault circuit interruptor, this will open the circuit when a difference in current is detected between the existing wires
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A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin
Mariana [72]

Answer:

The velocity of block = 0.188 \frac{m}{s}

Explanation:

Mass m = 5.6 kg

k = 1040 \frac{N}{m}

\mu = 0.26

x_{1} = 0.035 m  , v_{1} = 0

x_{2} = 0.02 m

From work energy theorem

K_{1} + U_{1} + W_{other} = K_{2} + U_{2}  --------- (1)

Kinetic energy

K = \frac{1}{2} k x^{2}  ------- (1)

Potential energy

U = \frac{1}{2} k x^{2} ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

R_{N} = mg

R_{N}  = 5.6 × 9.81 = 54.9 N

Friction  force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

W_{f} = - 21.9 × 0.015

W_{f} = - 0.329 J

Now kinetic energy

At point 1

K_{1} = \frac{1}{2} m v^{2} _{1}

K_{1} = 0

U_{1} = \frac{1}{2} k x^{2}

U_{1} = \frac{1}{2}  (1040) 0.035^{2}

U_{1} = 0.637 J

At point 2

K_{2} = \frac{1}{2} (5.6) v^{2} _{2}

K_{2} = 2.8 v_{2} ^{2}

Potential energy

U_{2} = \frac{1}{2}  k x_2^{2}

U_{2} = \frac{1}{2}  (1040) 0.02^{2}

U_{2} = 0.208 J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 v_{2} ^{2} + 0.208

2.8 v_{2} ^{2} = 0.1

v_{2} = 0.188 \frac{m}{s}

This is the velocity of block.

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4 years ago
Which of the following BEST describes the term "air resistance"?
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8 0
3 years ago
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Several different compounds can share the same empirical formula. The Molecular weight for three compounds with the empirical fo
Pie

Molecular formulas:

  1. CH₂O;
  2. C₂H₄O₂;
  3. C₆H₁₂O₆.
<h3>Explanation</h3>

The empirical formula of a compound tells only the ratio between atoms of each element. The empirical formula CH₂O indicates that in this compound,

  • for each C atom, there are
  • two H atoms, and
  • one O atom.

The molecular weight (molar mass) of the molecule depends on how many such sets of atoms in each molecule. The empirical formula doesn't tell anything about that number.

It's possible to <em>add</em> more of those sets of atoms to a molecular formula to increase its molar mass. For every extra set of those atoms added, the molar mass increase by the mass of that set of atoms. The mass of one mole of C atoms, two mole of H atoms, and one mole of O atoms is 12.0 + 2\times 1.0 + 16.0 = 30.0\;\text{g}.

  • CH₂O- 30.0 g/mol;
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  • C₃H₆O₃- 30.0 + 30.0 + 30.0 = 3 × 30.0 = 90.0 g/mol.

It takes one set of those atoms to achieve a molar mass of 30.0 g/mol. Hence the molecular formula CH₂O.

It takes two sets of those atoms to achieve a molar mass of 60.0 g/mol. Hence the molecular formula C₂H₄O₂.

It takes \dfrac{180.0}{30.0} = 6 sets of those atoms to achieve a molar mass of 180.0 g/mol. Hence the molecular formula C₆H₁₂O₆.

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3 years ago
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