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143°C = _____
-130 K
0 K
143 K
416 K **Weegy:** 143°C = 416 K. (More)

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Asked 9/20/2012 7:29:56 AM

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100 K = _____
-173°C
0°C
100°C
373°C **Weegy:** 100 K = -173°C (More)

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Updated 209 days ago|4/28/2014 10:17:01 AM

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The freezing temperature of water is zero on both the Kelvin and Celsius scales.
True
False **Weegy:** No. (More)

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Updated 9/21/2012 9:52:18 AM

4 Answers/Comments

This statement is False.

At the freezing point of water, the temperature of the Kelvin scale reads 273 K.

0º is the freezing point of water in Celsius.

At the freezing point of water, the temperature of the Kelvin scale reads 273 K.

0º is the freezing point of water in Celsius.

Added 9/20/2012 7:55:13 AM

0 °C = 273.15 K.

This definition fixes the magnitude of both the degree Celsius and the kelvin as precisely 1 part in 273.16 (approximately 0.00366) of the difference between absolute zero and the triple point of water. Thus, it sets the magnitude of one degree Celsius and that of one kelvin as exactly the same. Additionally, it establishes the difference between the two scales' null points as being precisely 273.15 degrees Celsius (-273.15 °C = 0 K and 0 °C = 273.15 K).[3]

This definition fixes the magnitude of both the degree Celsius and the kelvin as precisely 1 part in 273.16 (approximately 0.00366) of the difference between absolute zero and the triple point of water. Thus, it sets the magnitude of one degree Celsius and that of one kelvin as exactly the same. Additionally, it establishes the difference between the two scales' null points as being precisely 273.15 degrees Celsius (-273.15 °C = 0 K and 0 °C = 273.15 K).[3]

Added 9/20/2012 8:01:42 AM

@kidcarbon- yes the equal each other but the question asks if they both read zero. They do not both read zero.

Added 9/20/2012 8:09:05 AM

it is false.for kelvin the freezing point it is 273k but for celsius is 0c

Added 9/21/2012 9:52:19 AM

This answer has been flagged as incorrect.

Flagged by andrewpallarca

p1*v1=p2*v2
p1=.97atm
v1=3.1L
p2=.61
v2=? **Weegy:** The T1 and T2 in Charles' law equations V1/T1 = V2/T2 are temperature1 and temperature2. (More)

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Asked 9/20/2012 7:36:21 AM

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A hydraulic system for a dentist's chair is designed to be able to lift 3,112 newtons. The surface area over which this force is applied is 707 cm2. What force must be applied to the activator end of the system if the button has a surface area of 3.14 cm2?
700,700 N
1.41 N
0.0724 N
13.8 N **Weegy:** 3112 : 0.707 dm^2 = x : 0.00314 dm^2
x = 13.8 N
area=58.8cm^2=.0058 m^2
pressure=20275 Pascals or 2 N/cm^2
So the units on your choices are incorrect. (More)

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Asked 9/19/2012 10:59:36 AM

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