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4 years ago

Which statement represents how charges interact?

Physics

2 answers:

mario62 [17]4 years ago

7 0

Answer:

option (D) is correct.

Explanation:

This is based on the property of charges.

Like charges repel each other while like charges attract each other .

It means two positively charged particles or two negatively charged particles are repelled each other, while one positive and another negative charged particles attract each other.

The force of attraction or repulsion is given by the Coulomb's law.

According to this law, the force between two charges is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

marishachu [46]4 years ago

6 0

I believe the correct answer from the choices listed above is option D. Positive charges are attracted to negative charges. This would be based from the statement that opposites attract while like charges repel. Hope this answers the question. Have a nice day.

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Which of the following is not a factor in whether a reaction will spontaneously occur? A. Entropy change of the system B. Enthal

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3 years ago

The current through a certain heater wire is found to be fairly independent of its temperature. If the current through the heate

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Explanation:

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4 years ago

Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and

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Answer:

The maximum shear stress in shaft AB, $T_{ABmax}$ is 15 MPa

The maximum shear stress in shaft CD, $T_{CDmax}$ is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by

J = $\pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}$

Therefore, we have

$J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}$

Where:

D $_{AB}$ = Diameter of shaft AB = 30 mm = 0.03 m

r $_{AB}$ = Radius of shaft AB = 15 mm = 0.015 m

∴ $J_{AB} = \pi \times \frac{0.03^4}{32} =\pi \times \frac{0.015^4}{2}$ = 7.95 × 10⁻⁸ m⁴

and

$J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}$

Where:

D $_{CD}$ = Diameter of shaft CD = 36 mm = 0.036 m

r $_{CD}$ = Radius of shaft CD = 18 mm = 0.018 m

Therefore,

$J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2}$ = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= $1.58 \times \frac{2\pi }{360}$ rad = 2.76 × 10⁻² rad.

That is $\phi_r$ = 2.76 × 10⁻² rad.

However $\phi_r = \phi_{C/D} - \phi_{B/A}$

Where:

$\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$ and

$\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}$

$T_{AB}$ and $T_{CD}$ = Torque on shaft AB and CD respectively

$T_{AB}$ = Required

$T_{CD}$ = 500 N·m

$L_{AB}$ and $L_{CD}$ = Length of shafts AB an CD respectively

$L_{AB}$ = 600 mm = 0.6 m

$L_{CD}$ = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

$\phi_r = \phi_{C/D} - \phi_{B/A}$ = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

2.76 × 10⁻² rad = $\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}$

= $\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}$

Therefore;

T $_{AB}$ = 79.54 N.m

Where T = $T_{AB}$ + T $_{CD}$ =

Therefore T $_{CD total }$ = 500 - 79.54 = 420.46 N·m

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$\tau_{CDmax}$ = $\frac{T_{CD}\times R_{CD}}{J_{CD}}$ = $\frac{420.46\times 0.018}{1.65\times 10^{-7}}$ = 45.9 MPa

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