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3 years ago

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Physics

1 answer:

Dmitrij [34]3 years ago

6 0

Answer:

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An electric flux of 147 N*m^2/C passes through a flat horizontal surface that has an area of 0.824 m^2. The flux is due to a uni

Tresset [83]

Answer:231.16 N/C

Explanation:

Given

Electric Flux $=147 N-m^2/C$

Area(A) $=0.824 m^2$

Given Field point above $31.6 ^{\circ}$

Therefore angle between Area vector Electric Field =90-31.6= $58.4^{\circ}$

We know that Flux is given by

$\phi =\vec{E}\cdot \vec{A}$

$\phi =EAcos\theta$

$147=E\times 0.824\times cos(58.4)$

E=231.16 N/C

4 0

3 years ago

A particle with charge 8 µC is located on the x-axis at the point −10 cm , and a second particle with charge 3 µC is placed on t

bixtya [17]

Answer:Force on -7 uC charge due to charge placed at x = - 10cm

now we will have

towards left

similarly force due to -5 uC charge placed at x = 6 cm

now we will have

towards left

Now net force on 7 uC charge is given as

towards left

Explanation:

6 0

3 years ago

What is the distance from the earth's center to a point outside the earth where the gravitational acceleration due to the earth

Crazy boy [7]

R2^ 2 / R1 ^2 = g1 / g2 = 38

<span>R2 = R1 x √38 = 6.1644* R1 </span>

<span>R2 = 6.1644 x 6378 000 = 39316632.5 m</span>

8 0

3 years ago

It is harmful to settle near the airport why

Gemiola [76]

Yes, plane accidents can happen near you, it can harm your property and cause a disaster.

4 0

2 years ago

Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

3 years ago

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