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dybincka [34]
3 years ago
9

What are the three ways that a person can manipulate light?

Physics
1 answer:
BARSIC [14]3 years ago
6 0

The three ways a person can manipulate light would be the following:, filter, and the time the photograph is taken

 

<span>1.    </span>Angle - <span>The </span>camera angle<span> <span>marks the specific location at which the movie </span></span>camera<span> <span>or video </span></span>camera<span> is placed to take a shot.</span>

<span>2.    </span>Filter - Camera<span> <span>lens </span></span>filters<span> <span>still have many uses in digital photography, and should be an important part of any photographer's </span></span>camera<span> bag.</span>

<span>3.    </span>Time the photograph is taken - The golden hour, sometimes called the "magic hour", is roughly the first hour of light after sunrise, and the last hour of light before sunset, although the exact duration varies between seasons. During these times the sun is low in the sky, producing a soft, diffused light which is much more flattering than the harsh midday sun that so many of us are used to shooting in.

 

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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A hair dryer draws 11 A when it is connected to 120 V. If electrical energy costs $ 0.09/kW·h, what is the cost of using the hai
Papessa [141]

Answer:

The cost of using the hair dryer for 15 minutes is \$3.6\bar 6

Explanation:

The parameters given in the question are;

The electric current drawn by the the air dryer, I = 11 A

The voltage to which the hair dryer is connected, V = 120 V

The duration of usage of the hair dryer = 15 minutes = 60 minutes /4 = 1 hour/4 = 0.25 hour

The electrical energy costs $0.09/kW·h

The power consumed by the hair dryer = I × V = 11 × 120 = 1320 Watts = 1.32 kW

The energy used by the hair dryer in 15 minutes (0.25 hour) = 1.32 × 0.25 0.33 kW·h

The energy used by the hair dryer in 15 minutes (0.25 hour) = 0.33 kW·h

The energy cost = $0.09/(kW·h)

Therefore;

The cost of using the hair dryer for 15 minutes (0.25 hour) = 0.33 kW·h/($0.09/(kW·h)) = $33/9 = $3 2/3 = $3.6\bar 6.

6 0
3 years ago
Problem 31:
blagie [28]

Answer:

Bi. Current in 15.4 Ω (R₁) is 7.14 A.

Bii. Current in 21.9 Ω (R₂) is 5.02 A.

Biii. Current in 11.7 Ω (R₃) is 9.40 A.

C. Total current in the circuit is 21.56 A.

Explanation:

Bi. Determination of the current in 15.4 Ω (R₁)

Voltage (V) = 110 V

Resistance (R₁) = 15.4 Ω

Current (I₁) =?

V = I₁R₁

110 = I₁ × 15.4

Divide both side by 15.4

I₁ = 110 / 15.4

I₁ = 7.14 A

Therefore, the current in 15.4 Ω (R₁) is 7.14 A.

Bii. Determination of the current in 21.9 Ω (R₂)

Voltage (V) = 110 V

Resistance (R₂) = 21.9 Ω

Current (I₂) =?

V = I₂R₂

110 = I₂ × 21.9

Divide both side by 21.9

I₂ = 110 / 21.9

I₂ = 5.02 A

Therefore, the current in 21.9 Ω (R₂) is 5.02 A

Biii. Determination of the current in 11.7 Ω (R₃)

Voltage (V) = 110 V

Resistance (R₃) = 11.7 Ω

Current (I₃) =?

V = I₃R₃

110 = I₃ × 11.7

Divide both side by 11.7

I₃ = 110 / 11.7

I₃ = 9.40 A

Therefore, the current in 11.7 Ω (R₃) is 9.40 A.

C. Determination of the total current.

Current 1 (I₁) = 7.14 A

Current 2 (I₂) = 5.02 A

Current 3 (I₃) = 9.40 A

Total current (Iₜ) =?

Iₜ = I₁ + I₂ + I₃

Iₜ = 7.14 + 5.02 + 9.40

Iₜ = 21.56 A

Therefore, the total current in the circuit is 21.56 A

7 0
3 years ago
An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t
scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

C = \frac{Q}{V}  = e_0\frac{A}{d}

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

V = \frac{Q}{e_0} *\frac{d}{A}

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V

So, if we double the distance between the plates, the potential difference will also be doubled.

6 0
3 years ago
List out any 4 objectives of education​
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6 0
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A boy pulls his toy on a smooth horizontal surface with a rope inclined at 60 degrees to the horizontal. If the effective force
lana66690 [7]
I'm thinking that you're supposed to divide. So you would divide 5 into 60 and get 12
6 0
3 years ago
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