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coldgirl [10]
3 years ago
13

.) 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:

a)

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

d)

Phase angle Ф=87.3°

e)

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
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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
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