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3 years ago

What is another principle an artist can follow to create the illusion of depth on a flat surface?

1 answer:

3 0

It can be both flat or it can be when you have new eyeglasses on and you look down it makes you think the ground looks like that but its not

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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

3 years ago

Describe a situation where you can be traveling at a low speed but have an extremely high velocity

Reil [10]

Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

at 500 miles per hour

Explanation:

3 0

3 years ago

Read 2 more answers

Describe mechanical energy in your own words

Vlad [161]

Answer:

motion......

movement.....

3 0

3 years ago

Read 2 more answers

A 4.0 cm × 4.2 cm rectangle lies in the xy-plane. You may want to review (Pages 664 - 668) . Part A What is the electric flux th

padilas [110]

Answer:

(A). The flux is 0.336 N.m²/C

(B). The flux is zero.

Explanation:

Given that,

Length = 4.2 cm

Width = 4.0 cm

Electric field $E=(150 i-200 k)\ N/C$

Area vector is perpendicular to xy plane

(A). We need to calculate the flux

Using formula of flux

$\phi=E\cdot A$

Where, E = electric field

A = area

Put the value into the formula

$\phi=(150 i-200 k)\times(4.2\times10^{-2}\times4.0\times10^{-2})k$

$\phi=-200\times4.2\times10^{-2}\times4.0\times10^{-2}$

$\phi=-0.336\ N.m^2/C$

(B). Given electric field

$E=(150i-200j)\ N/C$

We need to calculate the flux

Using formula of flux

$\phi=E\cdot A$

Put the value into the formula

$\phi=(150 i-200 j)\times(4.2\times10^{-2}\times4.0\times10^{-2})k$

Here, The component of k is not given

So, the flux is

$\phi=0$

Hence, (A). The flux is -0.336 N.m²/C

(B). The flux is zero.

7 0

3 years ago

Read 2 more answers

a care starting from rest has an acceleration 0.3 m/s square, calculate the velocity and distance travelled by this car after 2

Anna [14]

Answer:

Final velocity (v) = 36 m/s

Distance traveled (s) = 2,160 m

Explanation:

Given:

Initial velocity (u) = 0

Acceleration (a) = 0.3 m/s

Time travel (t) = 2 minutes = 120 seconds

Find:

Final velocity (v) = ?

Distance traveled (s) = ?

Computation:

v = u + at

v = 0 + 0.3(120)

v = 0.3(120)

v = 36 m/s

Final velocity (v) = 36 m/s

Distance traveled (s) = ut + (1/2)at²

Distance traveled (s) = (0.5)(0.3 × 120 × 120)

Distance traveled (s) = 2,160 m

3 0

3 years ago