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OlgaM077 [116]
3 years ago
7

Liz has designed a sailboat. She wants to build a prototype of the boat. Liz has $12,000 to spend on the boat. $7,000 of that mo

ney is budgeted for wood. Liz needs 500 ft of wood planks.
Assuming that Liz will use only one type of wood for her entire project, which of the following could she use to stay on budget?

A. Marine plywood only.
B. She could use any of these woods listed.
C. Marine plywood of Douglas for only.
D. Marine plywood, Douglas fir, or white oak only

Physics
2 answers:
serious [3.7K]3 years ago
7 0
It would be marine plywood only
Naddika [18.5K]3 years ago
3 0

Answer:

A. Marine plywood only.

Explanation:

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A battery is connected to a light bulb that has 3 of resistance . If there is 0.5 A of current flowing what is the voltage on th
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1.5 V

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8 0
3 years ago
Find A and effective resistance.<br> (Ill give Brainliest if you provide explaination)
Alex777 [14]

Answer:

A = 2.4 A

R_{eq}  = 5 \ \Omega

Explanation:

The voltage in the circuit, V = 12 V

The given circuit shows four resistors with R₁ and R₂ arranged in series with both in parallel to R₃ and R₄ which are is series to each other

R₁ = 4 Ω

R₂ = 6 Ω

R₄ = 5 Ω

The voltage across R₃ = 6 V

Voltage across parallel resistors are equal, therefore;

The total voltage across R₃ and R₄ = 12 V

The total voltage across R₁ and R₂ = 12 V

The voltage across R₃ + The voltage across R₄ = 12 V

∴ The voltage across R₄ = 12 V - 6 V = 6 V

The current flowing through  R₄ = 6V/(5 Ω) = 1.2 A

The current flowing through R₃ = The current flowing through R₄ = 1.2 A

The resistor, R₃ = 6 V/1.2 A = 5 Ω

Therefore, we have;

The sum of resistors in series are R₁ + R₂ and R₃ + R₄, which gives;

R_{series \, 1} = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω

R_{series \, 2} = R₃ + R₄ = 5 Ω + 5 Ω = 10 Ω

The sum of the resistors in parallel is given as follows;

\dfrac{1}{R_{eq}}  = \dfrac{1}{R_{series \, 1}} + \dfrac{1}{R_{series \, 2}} = \dfrac{R_{series \, 2} + R_{series \, 1}}{R_{series \, 1} \times R_{series \, 2}}

Therefore;

R_{eq} =  \dfrac{R_{series \, 1} \times  R_{series \, 2}}{R_{series \, 1} + R_{series \, 2}}

Therefore;

R_{eq} =  \dfrac{10\times  10}{10 + 10} \ \Omega = 5 \ \Omega

R_{eq}  = 5 \ \Omega

The value of the current, <em>A</em>, in the circuit, I = V/R_{eq}

A = I = 12 V/(5 Ω) = 2.4 A

A = 2.4 A

3 0
3 years ago
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