Ask question

Ask question

All categories

2 years ago

7

Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on

each? A. 4.0x10-6 C 4.0x10-6 C B. 7.4x10-6 C 0.6x10-6 C C. 6.6x10-6 C 1.4x10-6 C D. 5.0x10-6 C 3.0x10-6 C

1 answer:

nexus9112 [7]2 years ago

8 0

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC (OR) q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

<u>B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C</u>

You might be interested in

A skateboarder with a mass of 45 kilograms is riding on a skateboard with a mass of 2.5 kilograms. What should be the velocity o

klasskru [66]

So momentum is just velocity times mass, this means Momentum = Velocity x Mass.
We can rearrange this to be Velocity = Momentum/Mass.

Since we know momentum and mass we can now solve.

Velocity = 264/(45+2.5)
= 5.56 m/s

5 0

2 years ago

Read 2 more answers

Where would the weight of an explorer be greater?The summit of Chimborazo, in Ecuador, which is at a distance of about 6,384 km6

FromTheMoon [43]

Answer:

The weight would be greater in the Mariana because the radius of the earth is lower.

Explanation:

We will make a comparison through constants and equations to see which one is more viable.

Using the force of gravity

$F = \frac{Gm}{R^2}$

F = Gravity force

G= Gravitational constant

m= mass

R is the radius of the earth, $R_E = 6384$ km and $R_M =6370$ km

Density $(\rho)$ is equal in both places,

-Gravity Force at Ecuador, $F_E=\frac{Gm}{R_E^2}$

-Gravity Force at bottom of Mariana trench, $F_M=\frac{Gm}{R_M^2}$

Making the relation,

$\frac{F_E}{F_M}= \frac{\frac{Gm}{R_E^2}}{\frac{Gm}{R_M^2}}$

$\frac{F_E}{F_M}= (\frac{R_M}{R_E})^2$

For the Ecuador,

$F_E= F_M * \frac{R_M^2}{R_E^2}$

If we take $F_M * R_M^2$ as a constant X then

$F_E= \frac{X}{R_E^2}$

So the gravity force of the place is inversely proportion of the radius

Same for the Mariana trench,

$FM= \frac{X}{R_M^2}$

Then, the weight would be greater in the Mariana because the radius of the earth is lower.

3 0

2 years ago

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a

grigory [225]

<h2>Answer:</h2>

(c) 5m/s²

<h2>Explanation:</h2>

Total acceleration (a) of a particle in a circular motion is the vector sum of the radial or centripetal acceleration ( $a_{C}$ ) of the particle and the tangential acceleration ( $a_{T}$ ) of the particle and its magnitude can be calculated as follows;

a = $\sqrt{(a_{C})^2 + (a_{T})^2}$ ---------------------(i)

<em>But;</em>

$a_{C}$ = $\frac{v^{2} }{r}$ ------------------------------(ii)

Where;

v = instantaneous velocity

r = radius of the circular path of motion

<em>From the question;</em>

v = 30m/s

r = 300m

(i) First let's calculate the centripetal acceleration by substituting the values above into equation (ii) as follows;

$a_{C}$ = $\frac{30^{2} }{300}$

$a_{C}$ = $\frac{900}{300}$

$a_{C}$ = 3m/s²

(ii) From the question, the velocity of the particle is increasing at a constant rate of 4m/s² and that is the tangential acceleration $a_{T}$ , of the particle. i.e;

$a_{T}$ = 4m/s²

(iii) Now substitute the values of $a_{C}$ and $a_{T}$ into equation (i) as follows;

a = $\sqrt{(3)^2 + (4)^2}$

a = $\sqrt{(9) + (16)}$

a = $\sqrt{25}$

a = 5m/s²

Therefore, the magnitude of its total acceleration a, is 5m/s²

5 0

2 years ago

What type of equilibrium is guaranteed by each condition of equilibrium

Ostrovityanka [42]

Answer: Conditions for equilibrium require that the sum of all external forces acting on the body is zero (first condition of equilibrium), and the sum of all external torques from external forces is zero (second condition of equilibrium). These two conditions must be simultaneously satisfied in equilibrium

Explanation: Hope this helped

5 0

2 years ago

A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the fi

allochka39001 [22]

Answer:

a. q2 = 16.4μC, positive charge

b. F = 0.900N

c. downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

$F_e=k\frac{q_1q_2}{r^2}$ (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

$q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C$

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

3 0

2 years ago