Answer: B. 12.25 W
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Explanation:
1) Force = Weight = 75 x 9.8 = 735 N
2.)Work = Force x Height = 735 x 5 = 3675 J
3.)Power = Work / Time = 3675 / (5 x 60) = 12.25 W
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Answer:
V2= 1.03L
Explanation:
Start off with what you are given.
V^1: 1.00L
T^1: 23°C
V^2?
T^2: 33°C
If you know your gas laws, you have to utilise a certain gas law called Charles' Law:
V^1/T^1 = V^2/T^2
Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.
(23+273 = 296) (33+273 = 306)
Multiply crisscross
1.00/296= V^2/306
296V^2 = 306
Dividing both sides by 296 to isolate V2, we get
306/296 = 1.0337837837837837837837837837838
V2= 1.03L
Answer:
(i) ΔU = 116 J
(ii) ΔU = 289 J
(iii) ΔU = 1 KJ
(iv) ΔU = 0 J
(v) ΔU = 3.25 KJ
Explanation:
first law:
(i) W = 153 J; Q = - 37 J ( Q ( - ), losing friction )
⇒ ΔU = 153 - 37 = 116 J
(ii) W = 289 J; Q = 0 ( insulated)
⇒ ΔU = W = 289 J
(iii) Q = 1 KJ , W = 0 ( isovolumetric process)
⇒ ΔU = Q = 1 KJ
(iv) isothermal ( constant temperature )
∴ ΔT = 0° ( isothermal )
⇒ ΔU = 0 J
(v) isobaric ( constant pressure )
⇒ ΔU = Q + W
∴ Q = 15.6 KJ
∴ W = - ∫ P dV = - P ΔV; W (-) the system performs a job and the volume increases
.
∴ P = 950 KPa * ( 1000 Pa / KPa ) = 950000 Pa = 950000 J/m³
∴ ΔV = 18 - 5 = 13 L * ( m³ / 1000 L ) = 0.013 m³
⇒ W = - ( 950000 J/m³) * ( 0.013 m³ ) = - 12350 J ( - 12.35 KJ )
⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )
⇒ ΔU = 3.25 KJ
