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Alisiya [41]
4 years ago
12

1. The illuminance on a surface is 6 lux and the surface is 4 meters from the light source. What is the intensity of the source?

Physics
2 answers:
liraira [26]4 years ago
8 0
1. Is A 96 candelas

2. I think it's D, I'm not sure

3. Is C 5 meters

Crank4 years ago
5 0

<u>Answer</u>

1) A. 96 Candelas

2) A. Both of these types of lenses have the ability to produce upright images.

3) C. 5 meters


<u>Explanation</u>

Q1

The formula for calculation the luminous intensity is;

Luminous intensity = illuminance × square radius

Lv = Ev × r²

= 6 × 4²

= 6 × 16

= 96 Candelabra

Q2

For converging lenses, an upright image is formed when the object is between the lens and the principal focus while a diverging lens always forms and upright image.

A. Both of these types of lenses have the ability to produce upright images.

Q3

Luminous intensity = illuminance × square radius

square radius = Luminous intensity/ illuminance

r² = 100/4

= 25

r = √25

= 5 m




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The work done by the electric force to move a 1 c charge from a to b is

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4 0
3 years ago
An antelope moving with constant acceleration covers the distance 68.0 m between two points in time 7.50 s. Its speed as it pass
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Answer:

A)The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope is 1.77 m/s²

Explanation:

The equations of the position and velocity of the antelope is given by the following expressions:

x = x0 + v0 · t + 1/2 · a ·t²

v = v0 + a · t

Where:

x = position of the antelope at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A) Let´s place the center of the frame of reference at the first point. The equation of position at t = 7.50 s will be:

x = x0 + v0 · t + 1/2 · a ·t²

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

We also know that at the second point the velocity is 15.7 m/s. Then at t = 7.50 the velocity will be 15.7 m/s.

v = v0 + a · t

15.7 m/s = v0 + a · 7.50 s

We can solve this equation for "a" and replace it in the equation of height to obtain "v0". Then:

a = (15.7 m/s - v0) / 7.50 s

Replacing it in the equation for position:

68.0 m = 0 m + v0 · 7.50 s + 1/2 · a · (7.50 s)²

68.0 m = v0 · 7.50 s + 1/2 · (15.7 m/s - v0) / 7.50 s · (7.50 s)²

68.0 m = v0 · 7.50 s + 7.85 m/s · 7.50 s - 3.75 s · v0

68.0 m - 7.85 m/s · 7.50 s = 3.75 s · v0

(68.0 m - 7.85 m/s · 7.50 s) / 3.75 s = v0

v0 = 2.43 m/s

The speed of the antelope at the first point is 2.43 m/s.

B) The acceleration of the antelope will be:

a = (15.7 m/s - v0) / 7.50 s

a = (15.7 m/s - 2.43 m/s) / 7.50 s

a = 1.77 m/s²

The acceleration of the antelope is 1.77 m/s²

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Draw a velocity-time diagram as shown below.

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The distance traveled during the acceleration phase is
s₁ = (1/2)at₁²
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