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Scilla [17]
2 years ago
13

If my candle has a brightness of 15 candela from a distance of 2 meter. What would the brightness be at a distance of 6 meters?

Physics
1 answer:
nata0808 [166]2 years ago
6 0

Answer:

5 candela

Explanation:

Let the candle brightness be B

Let the distance be d

The candle brightness is inversely proportional to the distance

B α 1/d

B = k/d

If my candle has a brightness of 15 candela from a distance of 2 meter, then;

15 = k/2

k = 30

To get the brightness when the distance is 6m;

B = k/d

B = 30/6

B = 5 candela

Hence the brightness will be 5 candela

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There's no such thing as one balanced force or one unbalanced force.

If ALL of the forces in a GROUP of forces acting on the same object
all add up to zero, then we say that the GROUP of forces is balanced.

If they don't, then the GROUP of forces is unbalanced.

Two or more forces can be balanced or unbalanced.
One force can't.

7 0
2 years ago
Which pair of elements has the most similar properties?
Vanyuwa [196]
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In the gene TATTCATTGTTA—TGATTT—ATTCG, CATTGTTA encodes for pepsin, a digestive enzyme. The rest of the sequence doesn’t code fo
MatroZZZ [7]

Answer:

TATTCATTCATTA—TGATTT—ATTCG

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Option TATTCATTCATTA—TGATTT—ATTCG is the correct answer. In the DNA segment which encodes pepsin, a purine base (G) guanine is replaced by another purine (A) adenine. This type of mutation is called a transition type point mutation.

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8 0
3 years ago
Which of the following is not a subatomic particle?
Fynjy0 [20]
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7 0
3 years ago
The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12 degree
lina2011 [118]
A) the periodic time is given by the equation;
 T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
   = 6.28 × √0.0673
    = 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
 thus; frequency = 0.614 Hz

b)  The vertical distance, the height is given by
 h= 0.66 cos 12
 h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
 mgh = 1/2mv²
  v² = 2gΔh = 2×9.81 × 0.01 
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v = 0.443 m/s

c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
   = 1/2 × 0.310 × 0.443²
   = 0.0304 J


4 0
3 years ago
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