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Lostsunrise [7]
2 years ago
11

PLEASE ANSWER THIS WORKSHEET ITS DUE TODAY WILL GIVE BRAINLIEST BALANCED AND UNBALANCED WORKSHEET

Physics
1 answer:
OLga [1]2 years ago
6 0

Answer:B. NET force: 2 resultant motion: left

ItsOniiSama avatar

C. Net force: 3 Resultant motion: Left

ItsOniiSama avatar

D. Net Force: 7 Resultant motion: right

ItsOniiSama avatar

E. Net Force:0 resultant motion: NO MOTION

ItsOniiSama avatar

F. NET Force: 3 resultant motion: Down

ItsOniiSama avatar

G. NET FORCE: 10 resultant motion: up

ItsOniiSama avatar

H. Net force: 3 Resultant motion: left

ItsOniiSama avatar

I. Net force: 50 Resultant motion: right

ItsOniiSama avatar

J. NET FORCE: 75 Resultant motion: down

ItsOniiSama avatar

K. Net force :200 Resultant motion: Right

ItsOniiSama avatar

L. Net force: 0 resultant motion:No motion

Explanation:

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The separation between two magnets is reduced by a factor of 5. How does the force between them change?
gtnhenbr [62]

Answer:Increase by a factor of  25

Explanation:

We know that magnetic force is proportional to \frac{1}{r^2}

where r is the distance between them

F\propto B\propto \frac{1}{r^2}

Thus F\propto \frac{1}{r^2}

Where B=magnetic field

If distance is reduced by a factor of 5 then force must increase by a factor of 25 because

r\rightarrow \frac{r}{5}

so F\propto \dfrac{1}{(\frac{r}{5})^2}

F\propto \frac{25}{r^2}

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2 years ago
Can someone tell me the acceleration?
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3 years ago
Explain the difference between impulse and impulsive force?
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7 0
3 years ago
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

#SPJ4

3 0
1 year ago
Light is an oscillating magnetic field alone. an oscillating electric field alone. electric and magnetic fields that oscillate p
cupoosta [38]

Explanation:

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