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antiseptic1488 [7]
4 years ago
11

Which is an example of a monthly fixed cost for a sandwich shop?

Physics
2 answers:
sveta [45]4 years ago
7 0
 i beleive it is A. rent
aivan3 [116]4 years ago
3 0
I think the correct answer from the choices listed above is option A. The rent is an<span> example of a monthly fixed cost for a sandwich shop. It is a fixed cost since you are required to pay for it per month. Hope this answers the question. Have a nice day.</span>
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Water is moving at a velocity of 2.00 m/s through a hose with an Internal diameter of 1.60 cm. What is the volume flow rate at t
Leya [2.2K]

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15 m/s

Explanation:

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3 years ago
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In the graph above, how does the acceleration at A compare with the acceleration at B?
abruzzese [7]
This is a speed/time graph.
The slope of the graph at each point is the time rate of change of speed
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That seems to be exactly what choice-c says.
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3 years ago
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Which equation represents a combustion reaction? Pb(NO3)2 + 2HCl → PbCl2 + 2HNO3 2SO2 + O2 → 2SO3 2C2H6 + 7O2 → 4CO2 + 6H2O Ca +
Naddik [55]

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2C2H6 + 7O2 → 4CO2 + 6H2O

Explanation:

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3 years ago
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A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi
Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

T = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}

7 0
4 years ago
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