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3 years ago

What is the electric flux through the surface a1 that encloses all three objects?

1 answer:

6 0

For the answer to the question above, I assume that the question is two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3,?two objects, O1 and O2 have charges +1.0 µC and -1.9 µC, respectively, and a third object, O3, is electrically neutral.
From Gauss's law:
Flux = ∫c E . dA = q/eo 

Since this surface encloses all 
charge, we can simplify: 

Flux = (q1+q2+q3)/eo 
Flux = 
( (1*10^-6)+(-1.9*10^-6)+(0) )/(8.85*10^-12) = -101694.92 N·m2/C

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A big lump of meat of mass 5Kg is hung from a spring balance in an elevator. Find the reading of the balance of (I) the elevator

Mrrafil [7]

The reading of the balance if ,

I ) If the elevator is moving with a steady speed = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N

T = m g + m a

T = Force

m = Mass

g = Acceleration due to gravity

a = Acceleration

m = 5 kg

g = 10 m / s²

I ) If the elevator is moving with a steady speed,

At steady speed, a = 0

T = ( 5 * 10 ) + ( 5 * 0 )

T = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s²,

a = 0.2 m / s²

T = ( 5 * 10 ) + ( 5 * 0.2 )

T = 50 + 1

T = 51 N

Therefore, the reading of the balance if ,

I ) If the elevator is moving with a steady speed = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N

To know more about reading on a spring balance

brainly.com/question/14819147

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11 months ago

A car is moving at a velocity of 25 km/h and increases velocity to 1200 km/h in 2 min. what is the acceleration?

OlgaM077 [116]

Answer:

a = 2.72 [m/s2]

Explanation:

To solve this problem we must use the following kinematics equation:

$v_{f} =v_{o} + a*t$

where:

Vf = final velocity = 1200 [km/h]

Vo = initial velocity = 25 [km/h]

t = time = 2 [min] = 2/60 = 0.0333 [h]

1200 = 25 + (a*0.0333)

a = 35250.35 [km/h2]

if we convert these units to units of meters per second squared

$35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]$

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In a oxidation-reduction reaction, the oxidizing agent always undergoes reduction.

Free_Kalibri [48]

In a oxidation-reduction reaction, the oxidizing agent always undergoes reduction.

TRUE.

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Read 2 more answers

Convert 1nanosecond in to its SI init

SOVA2 [1]

Convert 1nanosecond in to its SI init

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