Internal and external combustion engines are two types of heat engines: they convert thermal energy into mechanical energy. The main difference between internal and external combustion engine is that in internal combustion engines, the working fluid burns inside the cylinder, whereas in external combustion engines, combustion takes place outside the cylinder and heat is then transferred to the working fluid.
Answer:
from O toWl Q partical is accelerating with constant magnitude. and from Q to P is decelerating with constant magnitude.
Explanation:
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Like charges repel, while opposite charges attract.
Answer:
A) ΔU = 3.9 × 10^(10) J
B) v = 8420.75 m/s
Explanation:
We are given;
Potential Difference; V = 1.3 × 10^(9) V
Charge; Q = 30 C
A) Formula for change in energy of transferred charge is given as;
ΔU = QV
Plugging in the relevant values gives;
ΔU = 30 × 1.3 × 10^(9)
ΔU = 3.9 × 10^(10) J
B) We are told that this energy gotten above is used to accelerate a 1100 kg car from rest.
This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.
Thus;
P.E = K.E
ΔU = ½mv²
Where v is final velocity.
Plugging in the relevant values;
3.9 × 10^(10) = ½ × 1100 × v²
v² = [7.8 × 10^(8)]/11
v² = 70909090.9090909
v = √70909090.9090909
v = 8420.75 m/s
Answer:
Explanation:
Let the specific heat of material be s
heat lost by material = m₁ s (T 1 - T ) , (T 1 - T ) is fall in temp , m₁ is mass of material
= .45 x s x (91 - 31.4 )
= 26.82 s
Heat gained by water
= m₂ cw (T2 - T )
1.3 x 4186 x ( 31.4 - 23 )
heat lost = heat gained
m₂ cw (T2 - T ) = m₁ s (T 1 - T )
1.3 x 4186 x ( 31.4 - 23 ) = .45 x s x (91 - 31.4 )
45711.12 = 26.82 s
s = 1704.36
