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3 years ago

How does the current values in and out of the battery compare to the sum of currents going through the light bulbs?

1 answer:

6 0

The current values in and out of the battery is going to be equal to the sum of the currents going through the light bulbs. In a parallel circuit, the current is divided as it enters different branches and light bulbs. However, the amount of current leaving the battery is always going to be the same as the amount of current entering the battery. Therefore, while current divides in a parallel circuit, the total amount of current always remains the same.

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Which two staments explain how a cell's parts help it get nutrients

Goryan [66]

Answer:

I think it's A and D.

6 0

3 years ago

The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o

N76 [4]

Frequency = (speed) / (wavelength)

Frequency = (331 m/s) / (0.6 m) = 551.7 Hz

3 0

3 years ago

A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the

blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

6 0

3 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

$F = \frac{KQ^2}{r^2}$

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

4 0

3 years ago

A 5 newton force and a 7 newton force act concurrently on a point. As the angle between the forces is increased from 0 to 180 th

Reika [66]

Answer:

The magnitude of the resultant decreases from A+B to A-B

Explanation:

The magnitude of the resultant of two vectors is given by

$R=\sqrt{A^2 +B^2 +2AB cos \theta}$

where

A is the magnitude of the first vector

B is the magnitude of the second vector

$\theta$ is the angle between the directions of the two vectors

In the formula, A and B are constant, so the behaviour depends only on the function $cos \theta$ . The value of $cos \theta$ are:

- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is

$R=\sqrt{A^2 +B^2+2AB}=\sqrt{(A+B)^2}=A+B$

- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is

$R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}$

- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is

$R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B$

4 0

3 years ago