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drek231 [11]
3 years ago
13

A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?

Physics
1 answer:
Burka [1]3 years ago
5 0

Answer:

the force of attraction between the two charges is 4.2 x 10⁹ N.

Explanation:

Given;

the magnitude of first charge, q₁ = 0.06 C

the magnitude of the second charge, q₂ = 0.07 C

distance between the two charges, r = 3 m

The force of attraction between the two charges is calculated as ;

F = \frac{Kq_1q_2}{r^2}

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/C²

F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N

Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.

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Answer:

The change in volume is 6.885\times 10^{- 5}\

Solution:

As per the question:

Coefficient of linear expansion of Copper, \alpha = 17\times 10^{- 6}\ K^{- 1}

Initial Temperature, T = 0^{\circ} = 273 K

Final Temperature, T' = 100^{\circ} = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = 30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}

V' = V(1 + \gamma \Delta T)

\gamma = 3\alpha

V' = V(1 + 3\alpha \Delta T)

where

V' = Final volume

V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))

\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\

7 0
3 years ago
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the
igomit [66]

Answer:

d = 68.5 x 10⁻⁶ m = 68.5 μm

Explanation:

The complete question is as follows:

An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is  1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?

The answer can be given by using the formula derived from Young's Double Slit Experiment:

y = \frac{\lambda L}{d}\\\\d  =\frac{\lambda L}{y}\\\\

where,

d = slit separation = ?

λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m

L = distance from screen (detector) = 1.7 m

y = distance between bright fringes = 15.7 mm = 0.0157 m

Therefore,

d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\

<u>d = 68.5 x 10⁻⁶ m = 68.5 μm</u>

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According to ohm's law what is the resistance of a light bulb if the applied voltage is 21 volts and the current is 3 amps?
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