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jarptica [38.1K]
3 years ago
6

Which of the following does not contribute to the potential energy of an object?

Physics
1 answer:
Ivahew [28]3 years ago
3 0

Answer:

A. Gravitational constant

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Which statements best describe the work of Marie and Pierre Curie? Check all that apply.
AleksAgata [21]

B. They came up with the term “radioactivity.”

C. They conducted experiments with uranium-containing minerals and pure uranium.

E. They discovered two new radioactive elements

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3 years ago
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Which object is I’m equilibrium?
IrinaK [193]

Equilibrium is a state of no motion. If donuts are on the table, they are in equilibrium. Literally anything that is NOT moving relative to the earth is in equilibrium. The chair is in equilibrium, the table is in equilibrium, the cookies are in equilibrium, and so on.

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3 years ago
A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant
Alla [95]

Answer:

a) the work (W) done during the process is -2043.25 kJ

b) the work (W) done during the process is -2418.96 kJ

Explanation:

Given the data in the question;

mass of water vapor m = 10 kg

initial pressure P₁ = 550 kPa

Initial temperature T₁ = 340 °C

steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4

from superheated steam table

specific volume v₁ = 0.5092 m³/kg

so the properties of steam at p₂ = 550 kPa, and dryness fraction

x = 0.4

specific volume v₂ = v_f + xv_{fg

v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )

v₂ = 0.1377 m³/kg

Now, work done during the process;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.1377 - 0.5092 )

W = 5500 × -0.3715

W = -2043.25 kJ

Therefore, the work (W) done during the process is -2043.25 kJ

( The negative, indicates work is done on the system )

b)

What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses

x₂ = 100% - 80% = 20% = 0.2

specific volume v₂ = v_f + x₂v_{fg

v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )

v₂ = 0.06939 m³/kg

Now, work done during the process will be;

W = mP₁( v₂ - v₁ )

W = 10 × 550( 0.06939 - 0.5092 )

W = 5500 × -0.43981

W = -2418.96 kJ

Therefore, the work (W) done during the process is -2418.96 kJ

3 0
3 years ago
When there are more waves passing through the reference point in a period of time, what wave characteristic also increase?
lesya [120]
Frequency increases.
3 0
4 years ago
A 12.0-V emf automobile battery has a terminal voltage of 16.0 V when being charged by a current of 10.0 A. (a) What is the batt
forsale [732]

Answer:

(a) the battery’s internal resistance is 0.4 Ω

(b) the dissipated power dissipated inside the battery is 40 W

(c) the  rate 0.096°C/min

Explanation:

Given information:

emf, ε = 12 V

voltage, V = 16 V

current, I = 10 A

mass, m = 20 kg

specific heat, c = 0.300 kcal/kg°C = 1255.8 J/kg°C

(a) What is the battery’s internal resistance?

to calculate the internal resistance, we can calculate by using the following formula

V = ε - Ir

where

V = voltage (V)

I = current (A)

r = internal resistance (Ω)

ε = emf (V)

since the battery is being charged, the current is negative, so

V = ε - (-I) r

V = ε + Ir

r = (V- ε)/I

 = (16 - 12)/10

 = 0.4 Ω

(b) What power is dissipated inside the battery?

to determine the dissipated power in the battery, use the following equation

P = I²r

where

P = power (W)

P = (10)² (0.4)

  = 40 W

(c) At what rate (in °C/min ) will its temperature increase if its mass is 20.0 kg and it has a specific heat of 0.300kcal/kg⋅°C , assuming no heat escapes

to find the rate of temperature increase by

Q = m c ΔT

where

Q = thermal energy (J)

c = specific heat (J/kg°C)

ΔT = the change of temperature

to find the energy, we use

E = Pt

the energy is converted in one minute

E = 40 x 60

  = 2400 J

thus,

2400 = 20 x 1255.8 x ΔT

ΔT = 2400/(20 x 1255.8)

     = 0.096°C/min

3 0
3 years ago
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