Answer:
Explanation:
According to conservation of energy
q = ΔE + w
q is heat given , ΔE is increase in internal energy and w is work done by the gas.
i )
q = 50 kJ ; w = 30 kJ
Substituting the values
50 kJ = ΔE + 30 kJ
ΔE = 20 kJ ; Hence ΔE is positive.
ii )
q = - 25 kJ ; w = 45 kJ
- 25 kJ = ΔE + 45 kJ
ΔE = - 70 kJ .
ΔE is negative. It is endothermic.
iii )
q = 12.5 kJ ; w = - 3.5 kJ
12.5 kJ = ΔE - 3.5 kJ
ΔE = 16 kJ
Hence ΔE is positive.
Answer:
Explanation:
The minimum distance for takeoff is:
The correct answer is C. I leave an attached image to have a clearer idea of the light process that uses the Light when entering the eye. When you have farsightedness, the image is formed behind the eye so the defective effect is generated.
What is sought with the lenses is to correct this condition and put the image back at the point of the retina.
Answer:
a) 6.26 m/s
b) 7.67 m/s
Explanation:
The potential energy at height h0 is initially ...
PE0 = mgh0
At height h1, the potential energy is ...
PE1 = mgh1
The difference in potential energy is converted to kinetic energy:
PE0 -PE1 = KE1 = (1/2)m(v1)^2
Solving for v1, we have ...
mg(h0 -h1) = (1/2)m(v1)^2
2g(h0 -h1) = (v1)^2
v1 = √(2g(h0 -h1))
__
a) When the body is 1 m high, its speed is ...
v = √(2(9.8)(3 -1)) ≈ 6.26 m/s . . . at 1 m high
__
b) When the body is 0 m high, its speed is ...
v = √(2(9.8)(3 -0)) ≈ 7.67 m/s . . . when it reaches the ground