a fixed mass of a n ideal gas is heated from 50 to 80C at a constant pressure at 1 atm and again at a constant pressure of 3 atm
1 answer:
Answer:
The energy required is same for both cases since specific heat capacity (Cp) does not vary with pressure.
Explanation:
Given;
initial temperature, t₁ = 50 °C
final temperature, t₂ = 80 °C
Change in temperature, ΔT =80 °C - 50 °C = 30 °C
Pressure for case 1 = 1 atm
Pressure for case 2 = 3 atm
Energy required in both cases is given;
where;
Cp is specific heat capacity, which varies only with temperature and not with pressure.
Therefore, the energy required is same for both cases since specific heat capacity (Cp) does not vary with pressure.
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The correct order is (in decreasing order of gravity strength)
Jupiter - Neptune - Venus - Mars
In fact, Wayne's weight on each planet is given by
where m is Wayne's mass, which is a constant value, and g is the gravity strength at the surface of the planet. Therefore, the Wayne's weight W on each planet is directly proportional to the gravity strength of that planet: so the planet with the strongest gravity is the one where Wayne's weight is the greatest (Jupiter, 333 pounds), followed by Neptune (159), Venus (128) and Mars (53).
Jupiter - Neptune - Venus - Mars
In fact, Wayne's weight on each planet is given by
where m is Wayne's mass, which is a constant value, and g is the gravity strength at the surface of the planet. Therefore, the Wayne's weight W on each planet is directly proportional to the gravity strength of that planet: so the planet with the strongest gravity is the one where Wayne's weight is the greatest (Jupiter, 333 pounds), followed by Neptune (159), Venus (128) and Mars (53).
Answer:395.6 m/s
Explanation:
Given
mass of bullet
mass of wood block
Length of string
Center of mass rises to an height of
initial velocity of bullet
let and
be the velocity of bullet and block after collision
Conserving momentum
-------------1
Now after the collision block rises to an height of 0.38 cm
Conserving Energy for block
kinetic energy of block at bottom=Gain in Potential Energy
substitute the value of in equation 1
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 =
Answer:
No, both the thermometers will give the different reading.
Explanation:
Given,
- Both thermometer has same ice point =
- Both thermometer has same steam point =
- Distance between the ice point and steam point in both the thermometer is same of 100 division,
All the data given in both the thermometers are same, but the material in the thermometer is different due to this the reading at 60^o C will differ in both the thermometer. Because the reading on both the thermometer is depended upon the thermal expansion of the material inside it, but both the materials are different. Due to this the rise of fluid in the thermometer, i,e,. the volume of the fluid material in the thermometer will depend upon the thermal expansion. Hence both the material alcohol and mercury have the different thermal expansion, therefore the rise of the fluid in the thermometer also differ in both the thermometer.