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We want to know if the forces currently acting on the block will cause it to accelerate (i.e. move down the ramp), and if so, how much force would need to be applied to the block to prevent this. We can test both of these things simultaneously.
A force diagram of the situation:
Where
#vecf_s# is the force of static friction,#vecn# is the normal force,#vecF_g# is the force of gravity, and#vecF_(gx)# and#vecF_(gy)# are the parallel and perpendicular components of the force of gravity, respectively.Note that I used the force of static friction, which we want to keep the object from moving. If this is not enough, and the object is accelerating under the given conditions, this is the force of kinetic friction. Whether you call this static or kinetic friction initially will not affect the calculations.
Next, we can put together statements for the parallel and perpendicular forces:
#sumF_x=vecF_(gx)-vecf_s=mveca_x#
#sumF_y=vecn-vecF_(gy)=mcancel(veca_y)=0# Note that I have defined down the ramp as positive , and so the parallel component of gravity is positive, while the perpendicular component is negative. This can be seen from the force diagram.
Note there is no movement in the vertical direction either way, and so the vertical acceleration is zero, setting the sum of the perpendicular forces equal to zero.
For the object to remain at rest and not slide down the ramp, the net horizontal acceleration must also be zero. We want to know what force, if any, is necessary to make the horizontal acceleration equal zero. We can call this other force
#sumF_x=vecF_(gx)-vecf_s-vecF=0# Note that
#vecF# is negative. We want it to oppose the motion, which would occur down the ramp (positive).
We can solve for this unknown force:
#vecF=vecF_(gx)-vecf_s#
If no external force is necessary, we should get a value of at most
We know the equation for friction:
From the sum of forces statement for perpendicular forces, we know that
From the force diagram, we can produce equations for the parallel and perpendicular components of the force of gravity. By the magic of geometry, the angle between the vector for the force of gravity (total) and the perpendicular component is the same as the angle of incline of the plane. We see that the parallel component of gravity is opposite the angle, so we will use
We know that the force of gravity is given by
#vecF_g=mg#
#vecF_(gx)=mgsin(theta)#
#vecF_(gy)=mgcos(theta)#
Therefore,
Substituting these equations back into the equation we derived for the unknown force:
#vecF=mgsin(theta)-mumgcos(theta)#
#=>vecF=mg(sin(theta)-mucos(theta))#
We can now plug in our known values to solve for
#vecF=(14kg)(9.8m/s^2)(sin(pi/12)-(1/5)cos(pi/12))#
There is no pleasant exact value (not a great angle), but we get
#vecF~~9.0N#
We can confirm that the object would be accelerating without the added force:
#vecF_x=vecF_(gx)-vecf=mveca_x#
#=>veca_x=(cancel(m)gsin(theta)-mucancel(m)gcos(theta))/cancel(m)#
#=>veca_x=g(sin(theta)-mucos(theta))#
#=>veca_x=(9.8m/s^2)(sin(pi/12)-(1/5)cos(pi/12))#
#=>veca_x~~0.64m/s^2#
We can also check our answer:
#sumF_x=vecF_(gx)-vecf-9.00496831N=mveca_x#
#=>veca_x=(mgsin(theta)-mumgcos(theta)-9.00496831N)/m#
#=>veca_x=gsin(theta)-mugcos(theta)-9.00496831N#
#=>veca_x=((14kg)(9.8m/s^2)sin(pi/12)-(1/5)(14)(9.8m/s^2)cos(pi/12)-9.00496831N)/(14kg)#
#=>veca_x=0#
Let
Total resistance seen by battery
Current in the bulb is given by the expression
Power dissipated in the bulb is
Inserting given values we get from (1)
Taking square root of both sides we get
To calculate
As there are infinite many resistors, there will still be infinite many resistors if we detach the first two resistors from the front of nodes
As such the network reduces to sum of two resistances 1. resistance
For infinite resistor network we have an equation
#R_e=r+(rxxR_e)/(r+R_e)#
#=>(R_e-r)=(rxxR_e)/(r+R_e)#
#=>(R_e-r)xx(r+R_e)=(rxxR_e)#
#=>(R_e^2-r^2)-(rxxR_e)=0#
Using (3) we get
#(24^2-r^2)-24r=0#
#=>r^2 +24r-24^2=0#
Solving the quadratic and choosing positive root as resistance can not be negative
-.-.-.-.-.-.-.
Quadratic equation can also be solved using inbuilt graphic utility
The way the question is worded is quite confusing, but if that's the way it's worded, here's how I'm interpreting this.
Since the definitions of the three thermodynamic functions given are:
#DeltaH_(H_2O(l)) = int_(T_1)^(T_2) C_P(l)dT#
#DeltaS_(H_2O(l)) = int_(T_1)^(T_2) (C_P(l))/TdT#
#DeltaU_(H_2O(l)) = DeltaH_(H_2O(l)) - Delta(PV)#
we can work from there.
It seems reasonable (though tedious) that the question wants us to calculate for
This would illustrate the definitions given above.
CHANGE IN ENTHALPY OF HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR
#DeltaH_(H_2O(l)) = int_("293 K")^("373 K") C_P(l)dT#
In this temperature range, we may assume that
#color(blue)(DeltaH_(H_2O(l))^(293 -> "373 K")) ~~ C_P(l)int_("293 K")^("373 K") dT#
#= ("75.36 J/mol"cdot"K")("373 - 293 K")#
#=# #color(blue)("6028.8 J/mol")#
Include the
Next:
#color(blue)(DeltaH_(H_2O(g))^(373 -> "520 K")) ~~ C_P(g)int_("373 K")^("520 K") dT#
#= ("36 J/mol"cdot"K")("520 - 373 K")#
(not#36000# ; the kilo was a typo.)
#=# #color(blue)("5292 J/mol")#
This gives a total
#bbcolor(blue)(nDeltaH) = n(DeltaH_(H_2O(l))^(293 -> "373 K") + DeltaH_"vap" + DeltaH_(H_2O(g))^(373 -> "520 K"))#
#= ("1 mol")(6028.8 + 40695.7 + 5292) = "52016.5 J"#
#=# #bbcolor(blue)("52.017 kJ")#
CHANGE IN ENTROPY OF HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR
Similarly, the change in entropy over a temperature range can be calculated by the same kind of approximation about
#color(blue)(DeltaS_(H_2O(l))^(293 -> "373 K")) ~~ C_P(l)int_("293 K")^("373 K") 1/TdT#
#= ("75.36 J/mol"cdot"K")ln(373/293)#
#=# #color(blue)("18.19 J/mol"cdot"K")#
At constant temperature and pressure, which is how it is at the phase equilibrium we establish for vaporization,
#color(blue)(DeltaS_"vap") = (DeltaH_"vap")/T_b = ("40695.7 J/mol")/("373 K")#
#=# #color(blue)("109.10 J/mol"cdot"K"#
Finally, a similar process for
#color(blue)(DeltaS_(H_2O(g))^(373 -> "520 K")) ~~ C_P(g)int_("373 K")^("520 K") 1/TdT#
#= ("36 J/mol"cdot"K")ln(520/373)#
#=# #color(blue)("11.96 J/mol"cdot"K")#
So, the total
#bbcolor(blue)(nDeltaS) = n(DeltaS_(H_2O(l))^(293 -> "373 K") + DeltaS_"vap" + DeltaS_(H_2O(g))^(373 -> "520 K"))#
#= ("1 mol")(18.19 + 109.10 + 11.96)# #"J/mol"cdot"K"#
#=# #bbcolor(blue)("139.25 J/K")#
CHANGE IN INTERNAL ENERGY FOR HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR
Recall that
For the heating process, we'd need the densities of water at both temperatures to get accurate molar volumes (which are necessary because liquids are fairly incompressible). I think we can assume that
But just to see...
#rho_("293 K") = "998.21 g/L"#
#rho_("373 K") = "958.37 g/L"#
#barV_("293 K") = M_m/rho = "g"/"mol" xx "L"/"g" = 18.015/998.21 = "0.018047 L/mol"#
#barV_("373 K") = M_m/rho = 18.015/958.37 = "0.018798 L/mol"#
So for
#DeltaH = DeltaU + Delta(PV) = DeltaU + PDeltaV# ,
(at constant pressure)
#color(blue)(nDeltaU_(H_2O(l))^(293 -> "373 K")) = nDeltaH_(H_2O(l))^(293 -> "373 K") - PDeltaV_(H_2O(l))^(293 -> "373 K")#
#=# #("1 mol")("6028.8 J/mol") - ("1.01325 bar")("0.000751 L")xx("8.314472 J")/("0.0831345 L"cdot"bar")#
#=# #color(blue)("6028.7 J") ~~ DeltaH_(H_2O(l))^(293 -> "373 K")#
Indeed, the
For the vaporization, a similar idea follows in that
#PDeltaV_((l)->(g)) ~~ PV_((g)) = nRT#
#= ("1 mol")("8.314472 J/mol"cdot"K")("373 K") = "3101.30 J"#
Therefore:
#color(blue)(nDeltaU_"vap") = nDeltaH_"vap" - PDeltaV_((l)->(g))#
#= ("1 mol")("40695.7 J/mol") - ("3101.30 J")#
#=# #color(blue)("37594.4 J")#
Finally, for the heating of the gas, again, we can assume ideality to get the change in volume of the water vapor. At constant
#PDeltaV = nRDeltaT#
#= ("1 mol")("8.314472 J/mol"cdot"K")("520 - 373 K")#
#=# #"1222.23 J"#
So:
#color(blue)(nDeltaU_(H_2O(g))^(373 -> "520 K")) = nDeltaH_(H_2O(g))^(373 -> "520 K") - PDeltaV#
#= ("1 mol")("5292 J/mol") - ("1222.23 J")#
#=# #color(blue)("4069.77 J")#
So, the overall
#bbcolor(blue)(nDeltaU) = n(DeltaU_(H_2O(l))^(293 -> "373 K") + DeltaU_"vap" + DeltaU_(H_2O(g))^(373 -> "520 K"))#
#= ("1 mol")(6028.7 + 37594.4 + 4069.77)# #"J/mol"#
#=# #bbcolor(blue)("47.693 kJ")#
The two mirrors are inclined at an angle
It is clear that for third reflection the ray must hit the mirror normally or make angle of reflection
As dotted line at
Incident ray at
Now in triangle
Following the definition of work written above we see that work done (W) by a constant force (
So mathematically
So vectorially
This means work is scalar product of two vectors quantities. So work is a scalar quantity.
TORQUE
When force is applied to rotate a body around an axis the magnitude of rotational effect caused by the force depends on three quantities (1) the distance of point of application of force from axis of rotation,the magnitude of radius vector (
Mathematically
Magnitude of Torque
So vectorially it is the cross product of two vectors
So torque is a vector quantity. The direction of torque is determined by the thumb rule as shown in above figure.
Refer to the figure above, we need to treat iron ring and gap in the ring separately.
Magnetic ring.
Let the current to produce required magnetic flux in ring be
Magnetizing force produced in the iron ring
where
We know that for all media except vacuum Flux density
Inserting given values we get
Magnetic Flux density
Inserting vales we get
Magnetic flux
The ratio of the total flux produced
This equals flux in the air gap.
Air gap.
Magnetic flux for air gap
To produce flux in the gap required Magnetizing force
Inserting various values we get
from (1) and (2)
Total current
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