Which situation would create a field like the one shown here?

The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?

postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

$Initial\ Distance = 2cm$ ---- $r_1$

$New\ Distance = 4cm$ --- $r_2$

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

$F_1 = \frac{kq_1q_2}{r_1^2}$ --- Coulomb's law

Substitute 2 for r1

$F_1 = \frac{kq_1q_2}{2^2}$

$F_1 = \frac{kq_1q_2}{4}$

The new force (F2) is

$F_2 = \frac{kq_1q_2}{r_2^2}$

Substitute 4 for r2

$F_2 = \frac{kq_1q_2}{4^2}$

$F_2 = \frac{kq_1q_2}{4*4}$

$F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}$

Substitute $F_1 = \frac{kq_1q_2}{4}$

$F_2 = \frac{1}{4}*F_1$

$F_2 = \frac{F_1}{4}$

The new force is 1/4 of the previous force.

3 0

2 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago

Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi

Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

P₁ = m v

P₁ = 0.0575 x 26.7

P₁ = 1.535 kg.m/s

b) momentum of ball after collision

P₂ = m v'

P₂ = 0.0575 x (-29.5)

P₂ = -1.696 kg.m/s

c) change in momentum

Δ P = P₂ - P₁

Δ P = -1.696 -1.535

Δ P = -3.231 kg.m/s

d) using conservation of momentum

initial speed of racket = 0 m/s

M u + m v = Mu' + m v

M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

0.331 u' = 3.232

u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

5 0

3 years ago

If the bonds in the reactants of Figure 7-3 contained 432 kJ of chemical energy and the bonds in the

Yuri [45]

Answer:

The energy change would be 46kJ

The energy would be absorbed

Explanation:

The energy change during a chemical reation, i.e. the reaction energy, is equal to the chemical energy stored in the bonds of the products less the chemical energy stored in the bonds of the reactants.

Hence:

Energy change = 478 kJ - 432kJ = 46kJ

The change is positive, this is, the chemical energy of the products is greater than the chemical energy of the reactants.

That corresponds to the second graph, where the level of the energy of the products in the graph is higher than the level of the energy of the reactants. Therefore, the conclusion is that the reaction absorbed energy and it is endothermic.

4 0

3 years ago

First to answer gets brainliest

marissa [1.9K]

Letter na that would be the answer

4 0

3 years ago