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kaheart [24]
3 years ago
15

A transformer for a laptop computer converts a 120-V input to a 24-V output. If the output current for the transformer is 1.8 A,

find the input current.
Physics
1 answer:
astra-53 [7]3 years ago
3 0
To find for the input current, we use ratio and proportion concept. We use the equation:

I(input)/ I(output) = V(output)/ V(input)

I(input) / 1.8 A = 24 V / 120 V
I(input) = 0.36 A

Hope this answers the question. Have a nice day. Feel free to ask more questions.
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<h3>Answer:</h3>

209.236 kg · m/s

<h3>General Formulas and Concepts:</h3>

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<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
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Momentum Formula: P = mv

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

m₁ = 87.2 kg

v₁ = 2.87 m/s

m₂ = 0.0520 kg

v₂ = 789 m/s

<u>Step 2: Find Momentums</u>

<em>Football Player</em>

  1. Substitute [MF]:                    P = (87.2 kg)(2.87 m/s)
  2. Multiply:                                P = 250.264 kg · m/s

<em>Bullet</em>

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  2. Multiply:                                P = 41.028 kg · m/s

<u>Step 3: Find difference</u>

  1. Define equation:                    P₁ - P₂
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What is the angle of deviation in a plane mirror at normal incidence?​
Tems11 [23]

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The deviation of a mirror is equal to twice the angle of incidence.The total angle between the straight-line path and the reflected ray is twice the angle of incidence. This is called the deviation of the light and measures the angle at which the light has strayed from its initial straight-line path.

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The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa.
scoundrel [369]

Answer:

108.37°C

Explanation:

P₁ = Initial pressure = 101 kPa

V₁ = Initial volume = 530 m³

T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K

P₂ = Final pressure = 101 kPa (because it is open to atmosphere)

V₂ = Final volume = 530 m³

P₁V₁ = n₁RT₁

⇒101×530 = n₁RT₁

⇒53530 J = n₁RT₁

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\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347

Dividing the first two equations we get

1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K

∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C

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