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ozzi
1 year ago
13

Tuesday

Physics
1 answer:
irinina [24]1 year ago
3 0

If a teacher had 48 red pens and the ratio of red to blue pens she owns is 6:1, she will have a total of 56 pens

Let the number of red pens be R and the number of blue pens be B.

R = 48

R / B = 6 / 1

48 / B = 6 / 1

B = 48 / 6

B = 8

Total number of pens = Number of red pens + Number of blue pens

Total number of pens = R + B

Total number of pens = 48 + 8

Total number of pens = 56 pens

Therefore, she has a total of 56 pens

To know more about ratio

brainly.com/question/361700

#SPJ9

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A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito
Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

c =  \frac{c1 \times c2}{c1 + c2}

=  \frac{2 \times 4}{2 + 4}

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

= 1.33 \times 10 {}^{ - 6}  \times 8

= 10.64 \times 10 {}^{ - 6}

charge on 2.0μf capacitor is

\frac{Qeq}{2 \times 10 {}^{ - 6} }

=  \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

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3 0
2 years ago
8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?
Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

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

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

q_1=q_2=+1C are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N

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3 years ago
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Answer:

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Explanation:

Use equation for work

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-----------------------

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