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Answer:

Satellite moves faster in orbit when it is close to the planet it orbits, and slower when it is farther away.

Explanation:

The satellite experiences two forces while in orbit

  1. Force due to gravity #F_g# of the planet
    #F_g=G(M_pxxm_s)/R_O^2#
    where, #M_p and m_s# are mass of the planet and mass of the satellite respectively; #G# is Universal gravitational constant and #R_O# is the radius of the orbit measured from the center of the planet.
    We also know that #R_O=R_p+h#, where #R_p# is the radius of planet and #h# is the height of the satellite above planet's surface.
  2. Net centrifugal force #F_C# due to its circular motion
    #F_C=(m_sv^2)/R_O#
    where #v# is the velicity of the satellite.

As the satellite-planet system is in equilibrium, equating both forces we get
#G(M_pxxm_s)/R_O^2=(m_sv^2)/R_O#
#=>G(M_p)/R_O=v^2#
#=>v^2prop1/R_O#

From above equation it is evident that when a satellite is moved to a larger radius/higher from the planet’s surface, #R_O# increases. To keep the system in equilibrium, the velocity must decrease.

Heights of satellites above earth and their velocities.

freemars.org

Answer:

The maximum height of the ball is #~~11m#.

Explanation:

Ignoring air resistance, friction, etc., this problem can be solved with energy conservation.

#E_(mech)=K+U#

Where #K# is kinetic energy and #U# is potential energy. Thus, when energy is conserved in a system it should follow that

#E_i=E_f#

#K_i+U_i=K_f+U_f#

From this relationship we can see that if potential energy, for example, were to decrease, kinetic energy would have to increase in order for energy to be conserved. The opposite is also true. The idea is that energy is being transformed between two forms (potential and kinetic) rather than any of it being lost of gained. When we have to include friction in problems, for example, energy is not conserved, because some of it is lost to heat (#E_(th)#).

Initially, assuming the spring is compressed to point where it's height is insignificant (so that we need not worry about gravitational potential energy of the ball beforehand), all energy is stored as spring potential energy in the spring. Because nothing is moving prior to the launch, there is no kinetic energy. Therefore, we have only potential energy to begin with.

#U_(sp)=U_f+K_f#

After the launch, as a the spring decompresses, the energy stored within it as spring potential energy is transferred to the ball, which is observed as the ball shoots upward. The spring potential energy has been transformed into kinetic energy. However, as the ball rises, it is gaining altitude, and therefore gaining gravitational potential energy. Because the potential energy of the ball increases, its kinetic energy must decrease, and we observe this in the form of the ball stopping at some point in the air before falling back to the earth. If not, it would continue to fly upwards forever!

Because we are concerned with the maximum altitude of the ball, what we're really asking is at what point all of the potential energy from the spring has been transferred into gravitational potential energy, which is given by #U_g=mgh#. When the ball reaches its maximum altitude and pauses momentarily before falling, it has no kinetic energy (remember projectile motion?), and therefore we ultimately have only gravitational potential energy.

#U_(sp)=U_g#

Spring potential energy, #U_(sp)# is given by #1/2k(Δs)^2#.

#1/2k(Δs)^2=mgh#

( tl;dr ) Rearranging to solve for #h#,

#h=(1/2k(Δs)^2)/(mg)#

Using our known values for mass, the spring constant, and the compression of the spring, we can now calculate a value for #h#.

#h=(1/2(18N/m)(4/3m)^2)/(0.150kg*9.8m/s^2)#

#h=10.88...m#

#h≈11m#

Answer:

The force field is not conservative; #curl(vecF)!=0#.

Explanation:

If #vecF# is a vector (force) field in #RR^3#, then the curl of #vecF# is the vector #curl(vecF)# and is written as #gradxxvecF#. The vector field #vecF# is conservative iff #curl(vecF)=0#. A curl of zero does not mean that the vector field is definitely conservative, but is the first "pre-requisite." Ultimately, a vector field must have a potential function such as that #vecF=gradf# to be conservative, but if the field has a nonzero curl, it definitely does not have an associated potential function and therefore definitely is not conservative.

As stated above, the curl is given by the cross product of the gradient of #vecF# (in the form #< P, Q, R >#) and itself.

lamar.edu

We have #vecF=< x+y, 2z-y, 2y-z ># where

#P=x+y#

#Q=2z-y#

#R=2y-z#

The curl of the vector field is then given as:

enter image source here

We take the cross product as we usually would, except we'll be taking partial derivatives each time we multiply by a partial differential.

For the #veci# component, we have #del/(dely)(2y-z)-del/(delz)(2z-y)#

#=>(2-2)=0#

For the #vecj# component, we have #del/(delx)(2y-z)-del/(delz)(x+y)#

#=>(0-0)=0#

(Remember that if we take the partial of a function with respect to some variable which is not present, the partial derivative is #0# as we treat all other variables as constants)

Good so far!

For the #veck#component, we have #del/(delx)(2z-y)-del/(dely)(x+y)#

#=>(0-1)=-1#

This gives a final answer of #curl(vecF)=<0,0,-1>!=0#

#:.# Vector field is not conservative

Answer:

I get #~~9.0N#

Explanation:

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:

enter image source here

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 #vecF# and rewrite our net parallel force statement:

#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 #0#. If not, it indicates that the object would accelerate under the given conditions (only force of gravity, normal force, and force of friction).

  • We know the equation for friction: #vecf=muvecn#

  • From the sum of forces statement for perpendicular forces, we know that #vecn=vecF_(gy)#

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 #sin(theta)# to find it. Similarly, we see that the perpendicular component of gravity is adjacent to the angle, so we will use #cos(theta)# to find it.

We know that the force of gravity is given by #vecF_g=mg#

#vecF_(gx)=mgsin(theta)#

#vecF_(gy)=mgcos(theta)#

Therefore, #vecn=mgcos(theta)#.

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#:

#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=9.00496831N#.

#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#

#:.# the object would accelerate down the ramp at #0.64m/s^2# under the original conditions.

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#

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 #"293 K" -> "373 K"#, through the vaporization process, and on to the heating process of #"373 K" -> "520 K"#, actually...

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 #C_P(l)# is approximately constant, so 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 #DeltaH_"vap"# of #color(blue)("40695.7 J/mol")# for the vaporization process.

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 #DeltaH# for the heating, vaporization, and further heating process as:

#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 #C_P(l)# and #C_P(g)#:

#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, #DeltaG = 0#, so that #DeltaH - TDeltaS = 0#. Therefore:

#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 #DeltaS# going from #"373 K"# to #"520 K"# would be:

#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 #DeltaS# for this heating process on #"1 mol"# of water would be:

#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 #DeltaH = DeltaU + Delta(PV) = DeltaU + PDeltaV + VDeltaP#. Well, we've been at constant pressure, so we don't have to worry about the #VDeltaP# term.

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 #DeltaU ~~ DeltaH# for this first part because #PDeltaV# is going to be small.

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 #"1 mol"# of liquid water, #DeltaV = 0.018798 - 0.018047 = "0.000751 L"#, and so, from:

#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 #PDeltaV# term hardly mattered for liquid water.

For the vaporization, a similar idea follows in that #DeltaV_((l)->(g)) ~~ V_((g))#. At high temperatures, we can treat water approximately as an ideal gas, and so:

#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 #P# and #n#:

#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 #DeltaU# is:

#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")#

drawn

The situation as described in the question has been shown in above figure in which the black spot to be viewed is at the point P

Here we will use the following formula for refraction at curve surface to calculate the shift.

FORMULA

#color(blue)(mu_r/v-mu_i/u=(mu_r-mu_i)/R.........[1])#

Where

#mu_i->"refractive index of the medium of incident ray"#

#mu_r->"refractive index of the medium of refracted ray"#

#u->"object distance"#

#u->"image distance"#

#R->"radius of curvature of the curved refracting interface "#

When viewed from right the refraction will occur in two curved interface.
1) From #"air" to "glass"# then

#mu_i=1, " "mu_r=n," "u=-2r," "R=-r," "v=v_1(say)#

Inserting these in [1] we get

#color(green)(n/v_1-1/(-2r)=(n-1)/-r)#

#color(green)(=>n/v_1+1/(2r)=-n/r+1/r)#

#color(green)(=>n/v_1=-n/r+1/(2r)=-(2n-1)/(2r)#

#color(green)(=>v_1=-(2nr)/(2n-1)" where " n>1#

2) From #"glass" to "air"# then

#mu_i=n, " "mu_r=1," "u=-r-(2nr)/(2n-1)=-(r(4n-1))/(2n-1),#
#R=-2r," "v=v_2(say)#

Inserting these in [1] we get

#color(green)(1/v_2-n/(-(r(4n-1))/(2n-1))=(1-n)/(-2r)#

#color(green)(=>1/v_2+(n(2n-1))/(r(4n-1))=(n-1)/(2r)#

#color(green)(=>1/v_2=(n-1)/(2r)-(n(2n-1))/(r(4n-1))#

#color(green)(=>1/v_2=(4n^2-5n+1-4n^2+2n)/(2r(4n-1))#

#color(green)(=>1/v_2=-(3n-1)/(2r(4n-1))#

#color(green)(=>v_2=-(2r(4n-1))/(3n-1)#

So finally Shift of point P when viewed from right will be

#color(red)(S_R=PB-abs(v_2)=3r-abs(v_2)=3r-(2r(4n-1))/(3n-1))#

#color(red)(=>S_R=r/(3n-1)(9n-3-8n+2)#

#color(red)(=>S_R=(r(n-1))/(3n-1))#

When viewed from left the refraction will occur in one curved interface.

From #"glass" to "air only"# then

#mu_i=n, " "mu_r=1," "u=-r," "R=-2r," "v=v_3say)#

Inserting these in [1] we get

#color(blue)(1/v_3-n/-r=(1-n)/(-2r)#

#color(blue)(=>1/v_3+n/r=-1/(2r)+n/(2r)#

#color(blue)(=>1/v_3=-1/(2r)+n/(2r)-n/r#

#color(blue)(=>1/v_3=-1/(2r)(1-n+2n)#

#color(blue)(=>v_3=-(2r)/(n+1)#

So finally Shift of point P when viewed from left will be

#color(red)(S_L=PD-abs(v_3)=r-abs(v_3)=r-(2r)/(n+1)=(r(n-1))/(n+1))#

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    Ryuu answered · 2 hours ago · in Pressure
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