Given expression for velocity of a particle by
#v=(t−2)\ ms^-1#
Comparing with kinematic expression
#v=u+at# ...............(1)
we get
#u=-2\ ms^-1and a=1\ ms^-2#
(1). Kinematic expression for displacement is
#s=s_0+ut+1/2at^2#
Inserting given conditions we get the expression for displacement
#r=r_0+(-2)t+1/2(1)t^2#
#r=r_0-2t+1/2t^2#
To find
#r(6)=11=r_0-2(6)+1/2(6)^2#
#=>11=r_0-12+18#
#=>r_0=5\ m#
#r=5-2t+1/2t^2# ........(2)
(2). From (2)
#r(8)=5-2xx8+1/2(8)^2#
#=>r(8)=5-16+32#
#=>r(8)=21\ m#
(3). From (2)
#53=5-2t+1/2t^2#
#=>t^2-4t-96=0#
Solving the quadratic by split the middle term we get
#t=-8 and 12# ,
Ignoring
#v(12)=-2+1xx12=10\ ms^-1#
Power input of the refrigerator is
So,in
Now,let the refrigerator will take the temperature of water to
Given,
So,using the equation,
We,get,
Now,heat energy required for conversion of
So,to take this
So,this entire amount of heat will be equals to
So,
we get,from the above equation,
Now,density of water is
So,volume of this much water will be
Now,
so,
I get
We use Newton's Law of Cooling, which states that
Solving the differential equation, we get
where
Plugging in the values, we need to find
We got:
When
We get:
The equation becomes:
Now, we can find
Decreasing by
So, the final equation is:
After
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.
where
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
Proved
Alternative method
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.
(1)
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
From which
To find angle C we can use The Sine Rule:
The angle the resultant makes with the horizontal is therefore
(2)
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))#
Differentiating
#((delA)/(delV))_T = ((delU)/(delV))_T - T((delS)/(delV))_T#
From
#((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...
GASES
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
#20^@ "C"# :
In that case,
#((delU)/(delV))_T = -P + (Talpha)/kappa#
Thus,
#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.
Please see below.
.
Let's take a look at an NTC thermistor circuit:
The input of
We know that the circuit has a fixed resistor
The output voltage
It is difficult to find the Voltage-Resistance graph of an NTC thermistor in books but we can develop it here. We will designate the horizontal axis as the
Furthermore, in order to obtain a reasonably clear scale, we will consider the two resistances in the equation in units of
The graph will be:
But since we are dealing with positive voltages only, we will concern ourselves with the right hand side of the graph where both
As the graph shows, when
As far as filament lamps go, here are some graphs:
As far as metallic conductors go, their voltage-current, current-resistance and voltage-resistance relationships are linear and their graphs are straight lines that go through the origin. They are called Ohmic conductors:
Hope this helps.
I got this; however I am not sure I interpreted the question correctly...
Consider the diagram:
where:
Let us divide our problem into a component along the
Let us start considering the initial vertical component of
we can consider the expression from Kinematics:
with:
we get:
so that:
Let us now use the relationship from Kinematics:
Let us now consider part b; to get
Along
where
So we end up with:
On the Moon
so that:
the charge -4
This force has shown as blue vector as symbolically.
the charge -4
This force has shown as green vector as symbolically.
Since vectors
F=
F=480+8.89=488.89 N