This is what I get
Voltage vs time relationship of a charging
#E_s#is the DC supply across #RC#circuit of time constant #tau#.
It is given that
We also know that charge held in a capacitor can be written as
Therefore (1) becomes
Total charge would be maximum at
Given condition is Charge on the capacitor must be
Let the desired time be
From (3) and (4) we get
Inserting various values in (1) we get
Let the initial velocity of the projectile be
To calculate the time of flight we use the kinematic expression
Now the kinetic energy of the projectile at the time of projection is provided by the potential energy of the compressed spring. We know that
#m#is mass of the projectile, #k#is the spring constant and #x#is the compression of the spring.
Equating both we get
Inserting various values we get
You lift any number of balls and let go. The moving ball(s) then collide(s) with the stationary ball(s) and causes an equal number of balls to move on the other side.
This demonstrates conservation of momentum, because the number of balls moving before the collision is equal to the number of balls moving after the collision.
Momentum is defined as:
Each ball has the same mass, so this equivalent to saying the mass of moving balls remains constant from collision to collision. That takes care of one term. What about velocity? Well, the velocity of the moving balls right before the collision and the velocity of the moving balls on the other side right after the collision are equal. This can be seen, because the moving balls after the collision will never surpass the starting height of the balls on pre-collision side.
So if mass and velocity are unchanging for the system right before and right after collision, then the momentum will necessarily be constant: hence, conserved.
Given expression for velocity of a particle by
Comparing with kinematic expression
#u=-2\ ms^-1and a=1\ ms^-2#
(1). Kinematic expression for displacement is
Inserting given conditions we get the expression for displacement
(2). From (2)
(3). From (2)
Solving the quadratic by split the middle term we get
#t=-8 and 12#,
Now,let the refrigerator will take the temperature of water to
So,using the equation,
Now,heat energy required for conversion of
So,to take this
So,this entire amount of heat will be equals to
we get,from the above equation,
Now,density of water is
So,volume of this much water will be
We use Newton's Law of Cooling, which states that
Solving the differential equation, we get
Plugging in the values, we need to find
The equation becomes:
Now, we can find
So, the final equation is:
So, the temperature after half an hour will be approximately
The difference in range will be the height of the level of water in the tank times
You really want to use Torricelli's Law here.
The range of the stream is the horizontal distance the stream lands away from the tank. For the halfway hole, I have labeled this
How long will it take a water droplet to reach the ground once it has left the hole? Since it's vertical velocity component is controlled by gravity, the time it takes to land on the ground,
For the halfway hole,
Now we need to figure out the horizontal distance that the drop will travel in that time. It will simply be its horizontal velocity times the time.
We can find
The problem statement asks us to determine
Let the radius of the great conical mound of height
Given that the weight of the finished mound is
Now for the sake of our calculation let us consider the center of the circular base of the the mound
It is obvious that the rate of decrease of radius of the conical mound with height will be given by
Hence at an arbitrary height
So the volume of an imaginary circular disk of infinitesimal thickness
So weight of this thin disk will be
So work done against gravitational pull for lifting this imaginary thin disk to a height of
The total work done
Considering the center of mass of the conical mound which is placed at
So work done to heap up uniform material found at ground level which is equivalent to lift the COM to a height of
The electric field at a particular point is the force experienced by a test charge of +1 Coulomb.
I will assign a charge of +1C at the location specified and find the net force that it experiences. The electric field is found by dividing that force by unit charge.
I will show 2 ways that you can do this.
Coulombs Law gives us:
Here are the forces acting:
From Pythagoras we get:
By inspection you can see that
Now we find the resultant of these two forces.
We have a S ide A ngle S ide triangle ABC so we can apply The Cosine Rule:
From the diagram you can see that
To find angle C we can use The Sine Rule:
The angle the resultant makes with the horizontal is therefore
We can resolve
Now we can apply Pythagoras:
As you can see there is close agreement between the 2 methods so thats all good.
#DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV = int_(V_1)^(V_2) -P + T((delP)/(delT))_VdV#
Now decide what gas law to use, or what
Well, from the total differential at constant temperature,
#dU = cancel(((delU)/(delT))_VdT)^(0) + ((delU)/(delV))_TdV#,
so by definition of integrals and derivatives,
#DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV# #" "bb((1))#
The natural variables are
#dA = -SdT - PdV# #" "bb((2))#
This is also related by the isothermal Helmholtz relation (similar to the isothermal Gibbs' relation):
#dA = dU - TdS# #" "bb((3))#
#((delA)/(delV))_T = ((delU)/(delV))_T - T((delS)/(delV))_T#
#((delA)/(delV))_T = -P#
and also from
#((delS)/(delV))_T = ((delP)/(delT))_V#
since the Helmholtz free energy is a state function and its cross-derivatives must be equal. Thus from
#-P = ((delU)/(delV))_T - T((delP)/(delT))_V#
or we thus go back to
#barul|stackrel(" ")(" "DeltaU = int_(V_1)^(V_2) ((delU)/(delV))_TdV = int_(V_1)^(V_2)-P + T((delP)/(delT))_VdV" ")|#
And what remains is to distinguish between the last term for gases, liquids and solids...
Use whatever gas law you want to find it. If for whatever reason your gas is ideal, then
#((delP)/(delT))_V = (nR)/V#
and that just means
#((delU)/(delV))_T = -P + (nRT)/V#
#= -P + P = 0#
which says that ideal gases have changes in internal energy as a function of only temperature. One would get
#color(blue)(DeltaU = int_(V_1)^(V_2) 0 dV = 0)#.
Not very interesting.
Of course, if your gas is not ideal, this isn't necessarily true.
LIQUIDS AND SOLIDS
These data are tabulated as coefficients of volumetric thermal expansion
#alpha = 1/V((delV)/(delT))_P#
#kappa = -1/V((delV)/(delP))_T#
#alpha/kappa = [ . . . ] = ((delP)/(delT))_V#
at VARIOUS temperatures for VARIOUS condensed phases. Some examples at
In that case,
#((delU)/(delV))_T = -P + (Talpha)/kappa#
#color(blue)(DeltaU) = int_(V_1)^(V_2) -P + (Talpha)/kappadV#
#= color(blue)((alphaTDeltaV)/kappa - int_(V_1)^(V_2) P(V)dV)#
Next, the pressure will vary a lot if the volume varies a little for a solid or liquid...
Hence, to find
Then, we'll need to measure how the volume changes (to high precision!) with the change in pressure to empirically find the pressure as a function of the volume for the given solid or liquid.