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jeka57 [31]
3 years ago
14

A 14-Ω coffee maker and a 14-Ω frying pan are connected in series across a 120-V source of voltage. A 20-Ω bread maker is also c

onnected across the 120-V source and is in parallel with the series combination. Find the total current supplied by the source of voltage.
Physics
1 answer:
Ann [662]3 years ago
8 0

Answer:

The total current supplied by the source of voltage = 10.29 A

Explanation:

We have a 14-Ω coffee maker and a 14-Ω frying pan are connected in series.

Effective resistance = 14 + 14 = 28Ω

Now we have 28Ω and 20Ω in parallel

Effective resistance

             R=\frac{28\times 20}{28+20}=11.67\Omega

So we have resistor with 11.67Ω in a 120 V source of voltage.

We have equation V = IR

Substituting

               120 = I x 11.67

                 I = 10.29 A

The total current supplied by the source of voltage = 10.29 A

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Explanation:

To find the answer to this, you have to use the 4th kinematic equation:

v^{2} = v^{2}_{0}  + 2ax

You plug into the equation to get:

0 = 15^{2} + 2a(14)

solve for a to get

-8.04 m/s2

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Which of the following statements about price elasticity of demand is​ true?
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Answer:

B. The elastic portion of a​ straight-line, downward-sloping demand curve corresponds to the segment above the midpoint.

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Elasticity measures the sensitivity of one variable to another. Specifically it is a figure that indicates the percentage variation that a variable will experience in response to a variation of another one percent.

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How are volcanoes distributed
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3 years ago
Part A What will be the equilibrium temperature when a 227 g block of copper at 283 °C is placed in a 155 g aluminum calorimeter
stellarik [79]

Answer:

T = 20.84°C

Explanation:

From the law of conservation of energy:

Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water

m_cC_c\Delta T_c = m_wC_w\Delta T_w + m_aC_a\Delta T_a

where,

m_c = mass of copper = 227 g

m_w = mass of water = 844 g

m_a = mass of aluminum = 155 g

C_c = specific heat capacity of calorimeter = 385 J/kg.°C

C_w = specific heat capacity of water = 4200 J/kg.°C

C_a = specific heat capacity of aluminum = 890 J/kg.°C

\Delta T_c = change in temperature of copper = 283°C - T

\Delta T_w = change in temperature of water = T - 14.6°C

\Delta T_a = change in temperature of aluminum = T - 14.6°C

T = equilibrium temperature = ?

Therefore,

(227\ g)(385\ J/kg.^oC)(283^oC-T)=(844\ g)(4200\ J/kg.^oC)(T-14.6^oC)+(155\ g)(890\ J/kg.^oC)(T-14.6^oC)\\\\24732785\ J - (87395\ J/^oC) T = (3544800\ J/^oC) T - 51754080\ J+ (137950\ J/^oC) T-2014070\ J\\\\24732785\ J +51754080\ J+2014070\ J = (3544800\ J/^oC) T+(137950\ J/^oC+(87395\ J/^oC) T\\\\78560935\ J = (3770145\ J/^oC) T\\\\T = \frac{78560935\ J}{3770145\ J/^oC}

<u>T = 20.84°C</u>

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