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3 years ago

5

If a student tried to lift a 350 N box upwards with a lever that is 75 cm long, how much force would they need to provide given

a fulcrum 20 cm away from the box. 1312.5 N. 93.3 N. 65.6N. 2333N.

2 answers:

Lana71 [14]3 years ago

7 0

Answer:

93.3 N

Explanation:

Applying the equation of moment

M = r * F

r = distance from the applied force to the object = 75 cm

F = force applied = X

Moment of force = load * distance from load = 350 * 20

back to equation

350 * 20 = X * 75

7000 = 75 X

X ( force applied/needed) = 7000 / 75 = 93.3 N

Radda [10]3 years ago

3 0

We will take moment about the fulcrum point

Taking, box 350 N at a distance of 20 cm

we can write the moment equation as below:

350 N x 20 cm - X x 75 cm = 0

X = 350 N x 20 cm /75 cm

= 93.33 N

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Complete Question

Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

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ii $Q_2 = 4.4604 *10^{-11} \ C$

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The voltage of the battery is $V = 12.0 \ V$

The plate area of each capacitor is $A = 5.30 \ cm^2 = 5.30 *10^{-4} \ m^2$

The separation between the plates is $d = 2.65 \ mm = 2.65 *10^{-3} \ m$

The permittivity of free space has a value $\epsilon_o = 8.85 *10^{-12} \ F/m$

The dielectric constant of the other material is $z = 2.10$

The capacitance of the first capacitor is mathematically represented as

$C_1 = \frac{\epsilon * A }{d }$

substituting values

$C_1 = \frac{8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

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substituting values

$Q_1 = 1.77 *10^{-12} * 12$

$Q_1 = 2.124 *10^{-11} \ C$

The capacitance of the second capacitor is mathematically represented as

$C_2 = \frac{ z * \epsilon * A }{d }$

substituting values

$C_1 = \frac{ 2.10 *8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }$

$C_1 = 3.717 *10^{-12} \ F$

The charge stored in the second capacitor is

$Q_2 = C_2 * V$

substituting values

$Q_2 = 3.717*10^{-12} * 12$

$Q_2 = 4.4604 *10^{-11} \ C$

Now the total charge stored in the parallel combination is mathematically represented as

$Q_{eq} = Q_1 + Q_2$

substituting values

$Q_{eq} = 4.4604 *10^{-11} + 2.124*10^{-11}$

$Q_{eq} = 6.5844 *10^{-11} \ C$

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