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3 years ago

.) If the charges attracting each other in the problem above have equal magnitude, what is the magnitude of each charge?

Engineering

1 answer:

Sedaia [141]3 years ago

8 0

Answer:

Not seeing any other information, the best answer I can give is 2m.

Explanation:

M = magnitude

You see, if they have an equal charge, and you add them, it'd be 2 * m, or 2m.

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Consider weather conditions for which the prevailing wind blows past the penthouse tower on a tall building. The tower length in

ANTONII [103]

Answer:

Please find attached file for complete answer solution and explanation of same question.

Explanation:

3 0

3 years ago

A block of mass 0.75 kg is suspended from a spring having a stiffness of 150 N/m. The block is displaced downwards from its equi

olganol [36]

Answer:

a)f=2.25 Hz

b)Time period T=.144 s

c)tex]V_{max}[/tex]=0.42 m/s

d)Phase angle Ф=87.3°

e) $a_{max}=6.0041 [tex]\frac{m}{s^2}$

Explanation:

Natural frequency

$\omega _n=\sqrt {\dfrac{K}{m}}$

$\omega _n=\sqrt {\dfrac{150}{0.75}}$

$\omega _n$ =14.14 rad/s

w=2πf

⇒f=2.25 Hz

b) Time period

$=\dfrac{2π}{\omega _n}$

T= $\frac{1}{f}$

Time period T=.144 s

c)Displacement equation

$x=Acos\omega _nt+Bsin\omega _nt$

Boundary condition

t=o,x=0.03 m

t=0,v=.02m/s , V= $\frac{dx}{dt}$

Now by using these above conditions

A=0.03,B=0.0014

x=0.03 cos14.14 t+0.0014 sin14.14 t

⇒x=0.03003sin(14.14t+87.3)

$V_{max}=\omega_n X_{max}$

$V_{max}=14.14\times 0.03003$ =0.42 m/s

Phase angle Ф=87.3°

Maximum acceleration

$a_{max}=(\omega _n )^2X_{max}$

$a_{max}=(14.14)^20.03003$ =6.0041 $\frac{m}{s^2}$

3 0

4 years ago

Read 2 more answers

Different between boring and turning?

BartSMP [9]

Boring is not interesting. Turning is a place where a road branches off another.

6 0

4 years ago

Read 2 more answers

1. In a deck of cards, Kings, Queens and Jacks are called the face cards. If 4 cards are

VLD [36.1K]

The answer would be a standard of 26 to 2 because I did the same quiz

4 0

3 years ago

A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of

larisa86 [58]

Answer:

The field strength needed is 0.625 T

Explanation:

Given;

angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s

area of the rectangular coil, A = L x B = 0.0611 x 0.05 = 0.003055 m²

number of tuns of the coil, N = 300 turns

peak emf = 24 V

The peak emf is given by;

emf₀ = NABω

B = (emf₀ ) / (NA ω)

B = (24) / (300 x 0.003055 x 41.893)

B = 0.625 T

Therefore, the field strength needed is 0.625 T

4 0

3 years ago