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3 years ago

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In an oscillating lc circuit, when 75.0% of the total energy is stored in the inductor's magnetic field, (a) what multiple of th

e maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor?

1 answer:

olga2289 [7]3 years ago

4 0

<span>(a) E = Â˝ QÂ˛/C, so .. (b) E(max) = Â˝LiÂ˛ (i=current), so .</span>

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A fat person weighing 80 Kg falls on a concrete floor from 2m. If whole of the mechanical energy is converted into heat energy,

sasho [114]

Answer:

The heat produced is 1568 J

Explanation:

The given parameters are;

The mass of the person, m = 80 kg

The height from which the person falls, h = 2 m

Mechanical Energy, ME = Potential Energy, PE + Kinetic Energy, KE

At the height, from where the person falls, the initial velocity of the person = 0 m/s

Therefore;

The initial kinetic energy, K.E. = 1/2·m·v² = 1/2 × 80 kg × (0 m/s)² = 0 J

From which we have;

The Mechanical Energy, M.E. = The initial Potential Energy, P.E. + 0 J

∴ The Mechanical Energy, M.E. = The initial Potential Energy, P.E.

The initial Potential Energy, P.E. = m·g·h

Where;

m = The mass of the person

g = The acceleration due to gravity ≈ 9.8 m/s²

∴ The initial Potential Energy, P.E. = 80 kg × 9.8 m/s² × 2 m = 1568 J

The Mechanical Energy, M.E. = The initial Potential Energy, P.E. = 1568 J

The whole mechanical energy is converted into heat energy, therefore, we have;

The Mechanical Energy, M.E. = The heat energy = 1568 J

The heat produced = The heat energy = 1568 J.

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2 years ago

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom ris

Andrew [12]

Explanation:

Let us assume that forces acting at point B are as follows.

$\sum F_{x} = 0$

$T + F_{AB} Sin 60$ = 0 ...... (1)

$\sum F_{y}$ = 0

$F_{AB} Cos 60 + W$ = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

$\sigma_{allow} = \frac{T}{A}$

T = $(20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}$

= 3925 kip

From equation (1), $F_{AB}Sin (60^{o})$ = -3925

$F_{AB} \times -0.304 8$ = -3925

$F_{AB}$ = 12877.29 kip

From equation (2), -12877.29 (Cos 60) + W = 0

$-12877.29 kip \times \frac{1}{2} + W$ = 0

W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

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3 years ago

seraphim [82]

Answer:

ᵒᵒᵒᵒᵏᵏ

ᵏᵏᵏ

Explanation:

ᵒᵒᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏ

ᵏᵒᵒᵒᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵐᵐ

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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

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

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3 years ago

A ___________ is how strong the push or pull is

KatRina [158]

The answer is force

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3 years ago