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erastova [34]
2 years ago
12

The amount of energy required to raise the temperature of one gram of water by one celsius degree is called a

Physics
1 answer:
Elena L [17]2 years ago
5 0

The amount of energy required to raise the temperature of one gram of water by one celsius degree is called a calorie.

<h3>How can you define calorie?</h3>

In physics, we can define it as a unit of measurement that defines the amount of energy to raise the temperature of 1 gram  of water by 1°C. Each material has its unique calorie, however by definition we use water as it is the lowest of all the elements.

Normally, calorimetry is directly linked with the thermal capacity that an element has to contain heat.

See more about calorie at brainly.com/question/22374134

#SPJ1

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Answer:

solids

Explanation:

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3 years ago
What current flow (I) is associated with an input voltage of 5.0V and resistors R1 = 1.5 kiloohms and R2 = 0.5 kiloohms? Calcula
Y_Kistochka [10]

Answer:

current in series is 2.50 mA

current in parallel is 13.51 mA

Explanation:

given data

voltage = 5 V

resistors R1 = 1.5 kilo ohms

resistors R2 = 0.5 kilo ohms

to given data

current flow

solution

current flow in series is express as here

current = voltage / resistor    .................1

put here all value  in equation 1

current = 5 / (1.5 + 0.5)

current = 5 / 2.0

so current = 2.50 mA

and

current flow in parallel is express as

current = voltage / resistor   ....................2

put here all value in equation  2

current = 5 / (1/ (1/1.5 + 1/0.5))

current = 5 / 0.37

so current = 13.31 mA

5 0
3 years ago
In general, fitness evaluations are meant to help you measure your physical fitness against those of similar age and gender
kirill [66]

Answer:

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Explanation:

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3 years ago
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Can you answer this math homework? Please!
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Using the count data and observational data you acquired, calculate the number of CFUs in the original sample
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Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

6 0
3 years ago
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