Featured 2 months ago

This is what I get

Voltage vs time relationship of a charging

#V(t)=E_s(1-e^(-t/(tau)))# ......(1)

where#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

#Q=CV# .

Therefore (1) becomes

#Q(t)=CE_s(1-e^(-t/(tau)))# ......(2)

Total charge would be maximum at

#Q_max=CE_s# ........(3)

Given condition is Charge on the capacitor must be

Let the desired time be

#15/16Q_max=CE_s(1-e^(-(xtau)/(tau)))# .......(4)

From (3) and (4) we get

#15/16=(1-e^(-x))#

Inserting various values in (1) we get

#V(t)=5.877xx15/16=5.510\ V#

Featured 2 months ago

Let the initial velocity of the projectile be

Horizontal motion.

Distance traveled

#:.ucos30^@t=4#

#=>sqrt3/2ut=4# .......(1)

Vertical motion.

To calculate the time of flight we use the kinematic expression

#s=ut+1/2at^2#

Taking

#-1.2=usin30^@t+1/2(-9.8)t^2#

#=>4.9t^2-0.5ut-1.2=0# ........(2)

Eliminating

Ignoring the

#u=8/(sqrt3xx0.846)=5.46\ ms^-1# ......(4)

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

#KE_"projectile"=1/2m u^2# and

#PE_"spring"=1/2kx^2#

where#m# is mass of the projectile,#k# is the spring constant and#x# is the compression of the spring.

Equating both we get

#1/2kx^2=1/2m u^2#

#=>k=m u^2/x^2#

Inserting various values we get

#=>k=0.01 u^2/(0.1)^2#

#=>k=u^2#

Using (4)

#k=29.8\ Nm^-1#

Featured 2 months ago

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.

Featured 2 months ago

Given expression for velocity of a particle by

#v=(tâˆ’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#

Featured 2 months ago

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,

Featured 1 month ago

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

Featured 3 weeks ago

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

Featured 1 month ago

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

Featured 1 month ago

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.

Featured 1 month ago

#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** **isothermal compressibility**

#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"# :

#alpha_(H_2O) = 2.07 xx 10^(-4) "K"^(-1)# #alpha_(Au) = 4.2 xx 10^(-5) "K"^(-1)# (because that's REAL useful, right?)#alpha_(EtOH) = 7.50 xx 10^(-4) "K"^(-1)# -
#alpha_(Pb) = 8.7 xx 10^(-5) "K"^(-1)# -
#kappa_(H_2O) = 4.60 xx 10^(-5) "bar"^(-1)# #kappa_(Au) = 5.77 xx 10^(-7) "bar"^(-1)# #kappa_(EtOH) = 1.10 xx 10^(-4) "bar"^(-1)# #kappa_(Pb) = 2.33 xx 10^(-6) "bar"^(-1)#

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.