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True [87]
3 years ago
15

Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k =

8.99 × 1012N/C (figure is in attached file)
Physics
1 answer:
Pavlova-9 [17]3 years ago
4 0

Answer:

The magnitude of the electric field is 8.99*10^{12}N/C in the r direction.

Explanation:

The equation of the electric field is given by:

|\vec{E}|=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant is 8.99 *10^{9}\: Nm^{2}C^{-2}
  • q is the charge
  • r is the distance from A to q

|\vec{E}|=8.99*10^{9}\frac{12.5}{0.11^{2}}    

\vec{E}=9.29*10^{12} \vac{r} \: N/C

Therefore, the magnitude of the electric field is 8.99*10^{12}N/C in the r direction.

I hope it helps you!

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A man walks 400 m in the direction 45° north of east. Represent this vector graphically by
Marrrta [24]

Answer:

Explanation:

Coordinate system is one that describe the location of an object in a given plane. It implies the use of axes (coordinates) and points.

Given that the man in the question walks 400 m due north of east. The cardinal points can be used in this case, with the north and east cardinals as the required axis.

scale = \frac{length on drawing}{original length}

         = \frac{10}{400}

         = \frac{1}{40}

scale = 1:40

This is a reduced scale which implies that 1 cm on the drawing is equal to 40 m on the original length.

The man's direction is 45^{o} north of east.

The graphical drawing of the vector is herewith attached to this answer.

5 0
3 years ago
8. A man takes one minute to lift 4 bags of sugar each of weight 50N
timofeeve [1]

Answer:

D

Explanation:

the formula is F=mgh

so now you can write it like

m= 4×50=200

200×1.5×10=30000

6 0
3 years ago
the figure shows an initially stationary block of mass m on a floor. A force of magnitude 0.500mg is then applied at upward angl
Cerrena [4.2K]

Answer:

(a) 1.054 m/s²

(b) 1.404 m/s²

Explanation:

0.5·m·g·cos(θ) - μs·m·g·(1 - sin(θ))  - μk·m·g·(1 - sin(θ))  = m·a

Which gives;

0.5·g·cos(θ) - μ·g·(1 - sin(θ)   = a

Where:

m = Mass of the of the block

μ = Coefficient of friction

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the block

θ = Angle of elevation of the block = 20°

Therefore;

0.5×9.81·cos(20°) - μs×9.81×(1 - sin(20°)  - μk×9.81×(1 - sin(20°) = a

(a) When the static friction μs = 0.610  and the dynamic friction μk = 0.500, we have;

0.5×9.81·cos(20°) - 0.610×9.81×(1 - sin(20°)  - 0.500×9.81×(1 - sin(20°) = 1.054 m/s²

(b) When the static friction μs = 0.400  and the dynamic friction μk = 0.300, we have;

0.5×9.81·cos(20°) - 0.400×9.81×(1 - sin(20°)  - 0.300×9.81×(1 - sin(20°) = 1.404 m/s².

3 0
3 years ago
Help!!!!!!!!!!!!!!!!!!
GREYUIT [131]
I think wind is anther answer
5 0
3 years ago
Which component of health-related fitness is developed by performing a wall sit?
baherus [9]

Answer:

The answer is Muscular endurance

Explanation:

It is this because your seeing how long your muscles can with stand. I answered by guessing lol.

4 0
3 years ago
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