Question 11 of 15 what has an atomic number of 1?

What are the 2 factors that increase the electric force between objects?

harkovskaia [24]

The two factors that affect the electric force are the charges and the separation between the objects

Explanation:

The magnitude of the electrostatic force between two objects is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges of the two objects

r is the separation between the two objects

Looking at the equation, we see that the electric force between objects depends on two factors:

The charges of the objects: the force is directly proportional to the product of the two charges, therefore the greater the charges, the stronger the force
The distance between the two objects: the force is inversely proportional to the square of the distance, therefore the smaller the distance between the objects, the stronger the force

Learn more about electric force:

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4 0

3 years ago

What is the period of 60.0 hz electrical power?

Anarel [89]

Answer:

0.017 s

Explanation:

The period of a periodic signal is defined as the reciprocal of the frequency:

$T=\frac{1}{f}$

where

T is the period

f is the frequency

For the electrical power, the frequency is

f = 60.0 Hz

Substituting into the previous equation, we find the period:

$T=\frac{1}{60.0 Hz}=0.017 s$

8 0

3 years ago

You hang a light in front of your house using an

Kaylis [27]

The magnitudes of the forces that the ropes must exert on the knot connecting are :

F₁ = 118 N
F₂ = 89.21 N
F₃ = 57.28 N

Given data :

Mass ( M ) = 12 kg

∅₂ = 63°

∅₃ = 45°

<h3>Determine the magnitudes of the forces exerted by the ropes on the connecting knot</h3><h3 />

a) Force exerted by the first rope = weight of rope

∴ F₁ = mg

= 12 * 9.81 ≈ 118 kg

b) Force exerted by the second rope

applying equilibrium condition of force in the vertical direction

F₂ sin∅₂ + F₃ sin∅₃ - mg = 0 ---- ( 1 )

where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 ) applying equilibrium condition of force in the horizontal direction

Back to equation ( 1 )

F₂ = [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]

= [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]

= 89.21 N

C ) Force exerted by the third rope

Applying equation ( 2 )

F₃ = ( F₂ cos∅₂ / cos∅₃ )

= ( 89.21 * cos 63 / cos 45 )

= 57.28 N

Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :

F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N

Learn more about static equilibrium : brainly.com/question/2952156

6 0

2 years ago

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

Viktor [21]

Answer:

A

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

B

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$