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Paha777 [63]
3 years ago
6

1.What is true about heat capacity and specific heat?

Physics
1 answer:
Aleks04 [339]3 years ago
6 0
1-c. 2-d. 3-a. and 4 is -b.
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1 Point
irakobra [83]

Answer:

SECOND LAW OF NEWTON

Explanation:

When the rocket fires the engines the gases leave at high speed and collide with the space station, transferring an impulse given by the expression

                I = F t = Δp

As we can see this expression is a form of Newton's second law

           F = m a

           a = dv / dt

           F = m dv / dt

           F dt = m dv

           p = mv

           F dt = dp

Therefore the station moves through the SECOND LAW OF NEWTON

7 0
2 years ago
What is power, and what is its relationship to voltage and amperage? (4 points)
tigry1 [53]

Answer:

The relationship between amperage, voltage, and power is that power equals the amperage quantity times the amount of voltage.

Explanation: brainliest pls

5 0
3 years ago
Read 2 more answers
Determine the amount of potential energy of a 4.2 kg book that is placed on a shelf with a height of 0.9 meters. Round your answ
swat32

Answer:

that is a correct answer maybe .

5 0
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>

8 0
2 years ago
An object has an acceleration of 18.0 m/s/s. If the net force
Xelga [282]

Answer:

to ejam is god oh yum2

Explanation:

4 0
3 years ago
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