A household refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of 800 kJ/h.
1 answer:
Answer:
1454.54 kJ/h or 0.404 kW
Explanation:
Given:
Removal of heat by the refrigerator = 800 kJ/h
Coefficient of performance, COP of the refrigerator = 2.2
given that the refrigerator runs one-fourth of the time only
therefore, refrigerator removes the heat, Q = = 3200 kJ/h
Now, the Power drawn by the refrigerator while running =
=
= 1454.54 kJ/h
Hence, the power drawn by the refrigerator while running is = 1454.54 kJ/h
or 0.404 kW (as 1 kW = 3600 kJ/h)
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Answer:
(a) momentum of photon is 1.205 x 10⁻²⁷ kgm/s
velocity of electron is 1323.88 m/s
momentum of the electron is 1.205 x 10⁻²⁷ kgm/s
(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s
velocity of electron is 1654.85 m/s
momentum of the electron is 1.506 x 10⁻²⁷ kgm/s
(c) The momentum of the photon is equal to the momentum of the electron
Explanation:
(a)
wavelength of green light, λ = 550 nm
momentum of photon is given by;
velocity of electron is given by;
momentum of the electron is given by;
p = mv
p = (9.1 x 10⁻³¹) (1323.88)
p = 1.205 x 10⁻²⁷ kgm/s
(b)
wavelength of red light, λ = 440 nm
momentum of photon is given by;
velocity of electron is given by;
momentum of the electron is given by;
p = mv
p = (9.1 x 10⁻³¹) (1654.85)
p = 1.506 x 10⁻²⁷ kgm/s
(c) The momentum of the photon is equal to the momentum of the electron.
Answer:
<em>a) 6738.27 J</em>
<em>b) 61.908 J</em>
<em>c) </em>
<em></em>
Explanation:
The complete question is
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.
moment of inertia is given as
=
where m is the mass of the flywheel,
and r is the radius of the flywheel
for the flywheel with radius 1.1 m
and mass 11 kg
moment of inertia will be
=
= 6.655 kg-m^2
The maximum speed of the flywheel = 35 m/s
we know that v = ωr
where v is the linear speed = 35 m/s
ω = angular speed
r = radius
therefore,
ω = v/r = 35/1.1 = 31.82 rad/s
maximum rotational energy of the flywheel will be
E = = 6.655 x
= <em>6738.27 J</em>
<em></em>
b) second flywheel has
radius = 2.8 m
mass = 16 kg
moment of inertia is
=
=
= 62.72 kg-m^2
According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels
for the first flywheel, rotational momentum = = 6.655 x 31.82 = 211.76 kg-m^2-rad/s
for their combination, the rotational momentum is
where the subscripts 1 and 2 indicates the values first and second flywheels
= (6.655 + 62.72)ω
where ω here is their final angular momentum together
==> 69.375ω
Equating the two rotational momenta, we have
211.76 = 69.375ω
ω = 211.76/69.375 = 3.05 rad/s
Therefore, the energy stored in the first flywheel in this situation is
E = = 6.655 x
= <em>61.908 J</em>
<em></em>
<em></em>
c) one third of the initial energy of the flywheel is
6738.27/3 = 2246.09 J
For the car, the kinetic energy =
where m is the mass of the car
is the velocity of the car
Equating the energy
2246.09 =
making m the subject of the formula
mass of the car m =