S=56, u=0, v=33, a=?, t=3.4
v=u+at
33=3.4 a
a = 9.7m/s^2
Answer:
Explanation:
Let c be the circumference and r be the radius
c = 2πr , r = c / 2π , area A = π r² = π (c/2π )² = (1/4π) x c²
flux (ψ) = BA = 1 X 1/4π X c²
dψ/dt = 1/4π x 2c dc/dt =1/2π x c x dc/dt
at t = 8 s
c = 161 - 13 x 8 = 57 cm , dc/dt = 13 cm/s
e = dψ/dt = (1 / 2π )x 57 x 13 x 10⁻⁴ = 118 x 10⁻⁴ V.
Answer:
(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%
Explanation:
Using the formula that relate heat and work from the thermodynamic theory as:
solving to Q_out we get:
this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: 
where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:
Here is the highly detailed, arcane, complex, technical form of Ohm's Law that is needed in order to answer this question ===> I = V / R .
Current = (voltage) / (resistance)
Current = (1.5 V) / (10 Ω)
<em>Current = 0.15 Ampere</em>
