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2 years ago

14

The distance of an illuminated object from its light source is decreased to 1/3 while the luminous flux remains constant. Which

is true of the illuminance on the object

2 answers:

Nostrana [21]2 years ago

6 0

Answer:

The illuninance being multiplied by 9

Explanation:

GaryK [48]2 years ago

6 0

Answer:

The illuminance will be multiplied by 9.

Explanation:

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What would be the distance moved if we had a 70 n force and work done is 8j

Lostsunrise [7]

Answer:

0.1143m

Explanation:

W=f×s

8=70s

make s the subject of the formula

s=8/70

=0.1143m

3 0

2 years ago

Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha

Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, $A_p = 180 cm^2$

- The charge on each plate, $q = 17 * 10^-^6 C$

- Permittivity of free space, $e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}$

- The radius for the flux region, $r = 3.3 cm$

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

$A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2$

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

σ = $\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\$

σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

$E+ = E- = \frac{sigma}{2*e_o} \\\\$

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

$E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\$

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

Φ = E_net * Ar * cos ( θ )

Φ = (106214689.26553) * (0.00342) * cos ( 5 )

Φ = 361872 N.m^2 / C

5 0

2 years ago

The concepts in this problem are similar to those in Multiple-Concept Example 4, except that the force doing the work in this pr

steposvetlana [31]

Answer:

Explanation:

Given that:

distance (z) = 7.86 m

mass of the person = 81.7 kg

Acceleration (a) = 0.729 m/s²

By using Newton's second law along the vertical axis:

$T - mg = ma\\ \\ T = ma + mg \\ \\ T = m(a+g) \\ \\$

T = 81.7 (0.729 +9.8)

T = 860.22 N

The work done now is:

$W_T = T\times z \\ \\$

$W_T = 860.22 \times 7.86$

$\mathbf{W_T =6761.3292\ N}$

5 0

3 years ago

What is the speed of a 50 g rock if its de broglie wavelength is 3.32x10^-34m?

ivann1987 [24]

We can find the momentum of the rock by using De Broglie's relationship:
$p= \frac{h}{\lambda}$
where
p is the momentum
h is the Planck constant
$\lambda$ is the De Broglie's wavelength

By using $\lambda=3.32 \cdot 10^{-34} m$ , we find
$p= \frac{6.6 \cdot 10^{-34} Js}{3.32 \cdot 10^{-34} m}=1.99 kg m/s$

The momentum of the rock is
$p=mv$
where $m=50 g=0.05 kg$ is the mass and v is its velocity. Rearranging the equation, we find the speed of the rock:
$v= \frac{p}{m}= \frac{1.99 kg m/s}{0.05 kg}=39.8 m/s$

7 0

3 years ago

Somebody please help me!! Timed test

Olegator [25]

Answer: i dont know sorry

Explanation:

3 0

3 years ago