Answer:
52.5°C
Explanation:
The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state
mh1 + W = mh2
h2 = h1 + W/m
h1 + Wα1/V1
242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg
=287.4 kJ/kg
From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation
i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)
= 50°C + 60 - 50/295.15 - 284.79
(287.4 - 284.79)°C
= 52.5°C
Answer:
The speed of the block is 4.96 m/s.
Explanation:
Given that.
Mass of block = 1.00 kg
Spring constant = 500 N/m
Position 
Coefficient of friction = 0.350
(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less
Using formula of kinetic energy and potential energy
Put the value into the formula
Hence, The speed of the block is 4.96 m/s.
Answer: it depends on the mass of the pendulum or on the size of the arc through which it swings.
Explanation:
Answer:
2.45 J
Explanation:
The following data were obtained from the question:
Mass (m) = 0.5 kg
Height (h) = 1 m
Kinetic energy (KE) =?
Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 1/2 = 0.5 m
Final velocity (v) =?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 0.5)
v² = 9.8
Take the square root of both side
v = √9.8
v = 3.13 m/s
Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:
Mass (m) = 0.5 kg
Velocity (v) = 3.13 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 3.13²
KE = 0.25 × 9.8
KE = 2.45 J
Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J
A- sulfur oxide
B- carbon monoxide
