A heat engine exhausts 8900 j of heat while performing 2800 j of useful work. part a what is the efficiency of this engine?

1 answer:

7 0

Given: Heat Qout means useful work = 2800 J

Heat Qin = 8900 J

Required; Efficiency = ?

Formula: Efficiency = Qout/Qin = x 100%

= 2800 J/8900 J = 0.3146 X 100 %

Efficiency = 31.46%

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A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

3 0

2 years ago

Which of the following examples best represents an object with balanced forces acting upon it?

kirill [66]

D-a is the answer can I get braillist

7 0

2 years ago

Read 2 more answers

How would the period of a simple pendulum be affected if it were located on the moon instead of the earth?

OlgaM077 [116]

Answer:

On moon time period will become 2.45 times of the time period on earth

Explanation:

Time period of simple pendulum is equal to $T=2\pi \sqrt{\frac{l}{g}}$ ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth