Part (2) of Caleb's solutions to the Physics Test:

6)(a) The equilibrium temperature is T(E) = ½ [T(A) + T(B)]

b) Initially T(A) > T(B) (from diagram)

The change in entropy is given by dS = Q/T

And the change for a cold body is such that:

S increases by Q/ T_c (in this case T_c = T(B)

And S decreases by Q/ T_h (in this case T_h = T(A))

So, B undergoes a positive change in entropy: dS = Q/ T(B)

c) And A undergoes a negative change in entropy: -dS = -Q/T(A)

d) The total change in entropy is: Q/ T(B) + (-Q/T(A))

6)(a) The equilibrium temperature is T(E) = ½ [T(A) + T(B)]

b) Initially T(A) > T(B) (from diagram)

The change in entropy is given by dS = Q/T

And the change for a cold body is such that:

S increases by Q/ T_c (in this case T_c = T(B)

And S decreases by Q/ T_h (in this case T_h = T(A))

So, B undergoes a positive change in entropy: dS = Q/ T(B)

c) And A undergoes a negative change in entropy: -dS = -Q/T(A)

d) The total change in entropy is: Q/ T(B) + (-Q/T(A))

= Q /½ [T(A) + T(B)]

7) a) The force on the charge q is just the electric force F = qE where E is the electric field strength, so F = (3 x 10^-9 C) (200 V/m) = 6 x 10^-7 N

(b) The work done omn the charge by the field is just the force (F) applied times the displacement (s)

So: W = Fs = (6 x 10^-7 N) (0.50m) = 3 x 10^-7 Joule

(c) The potential difference [V(a) - V(b)] = E s

= (200 V/m) (0.5m) = 100 Volts

(8) Caleb wrote:

---

A sketch of the situation is shown:

The solution can be gleaned if one reasons here that the kinetic energy of the electron (say in the electron beam) is equal to the energy acuired by being accelerated through the voltage V (20,000V). Thus:

a) ½ mv^2 = qV = eV

where q = e is the electronic charge

Then the velocity of a given electron in the beam will be:

v = [2 eV/m]^1/2

but note here the quantity (e/m) is just the electron charge to mass ratio which can be obtained from a table of fundamental constants as: (e/m) = 1.8 x 10^11 C/kg

therefore, v = [ 2 (1.8 x 10^11 C/kg) (20,000 V)]^1/2 = 8.5 x 10^7 m/s

To get the deflection of the beam (y) we need to find the time taken to reach the screen first.

t = x/ v = 0.4m / (8.5 x 10^-7 m/s) = 4.7 x 10^-9 s

Then the deflection can be obtained using the basic kinematic form:

y = ½ at^2

where a is the acceleration but incepted by the effect of the Earth's magnetic field B.

For a given magnetic field, the force exerted is F(B) = qvB = evB

In this case:

F(B) = (1.6 x 10^-19 C) (8.5 x 10^7 m/s) (5 x 10^-5 T) = 6.8 x 10^-16 N

This is the force applied by the Earth's magnetic field to the electrons in the beam

By Newton's 2nd law: F = ma

ma = e vB so the acceleration a = e vB/ m = (e/m) v B

where (e/m) is the charge to mass ratio already obtained earlier, so:

a = (1.8 x 10^11 C/kg) ((8.5 x 10^7 m/s) (5 x 10^-5 T) = 7.6 x 10^14 m/s^2

Then, the displacement is found from:

y = ½ at^2 = y = ½ ( 7.6 x 10^14 m/s^2) ( 4.7 x 10^-9 s) = 8.4 x 10^-3 m

or 8.4 mm

(9) We have for force balance:

F = mg = u(o) I1 (I2) L/ 2 pi r

Where u(o) is the magnetic permeability = 4 pi x 10^-7 H/m

Since the same current runs through parallel wires on the balance, I1 = I2 = I

so: I1(I2) = I^2

and we can solve for the current needed to balance the force (mg) = 10N

I = [ 2 pi r (mg)/ u(o) L]^1/2

I = [ 2 pi (0.1m) (10N)/ (4 pi x 10^-7 H/m)(1m)]^1/2

and I = 2.2 x 10^3 A

10) The new half life curve, for T( ½ ) = 250,000 years is shown

After three million years, how much of the original sample will be left?

After one half life (250 ky) it is ½ gram so this is reduced by halves over successive half lives

Then we have:

After 2T( ½ ): 1/4 g

After 3T( ½ ): 1/8 g

After 4T( ½ ): 1/16 g (One million years)

After 5T( ½ ): 1/32 g

After 6T( ½ ): 1/64 g

After 7T( ½ ): 1/128 g

After 8T( ½ ): 1/256 g (Two million years is eight half lives)

After 9T( ½ ): 1/512 g

After 10T( ½ ): 1/1024 g

After 11T( ½ ): 1/2048 g

After 12T( ½ ): 1/ 4096 g

so the amount left after 3 million years = 1/ 4096 g

= 2.4 x 10^-4 g or 0.24 mg (millgram)

7) a) The force on the charge q is just the electric force F = qE where E is the electric field strength, so F = (3 x 10^-9 C) (200 V/m) = 6 x 10^-7 N

(b) The work done omn the charge by the field is just the force (F) applied times the displacement (s)

So: W = Fs = (6 x 10^-7 N) (0.50m) = 3 x 10^-7 Joule

(c) The potential difference [V(a) - V(b)] = E s

= (200 V/m) (0.5m) = 100 Volts

(8) Caleb wrote:

*?*---

A sketch of the situation is shown:

The solution can be gleaned if one reasons here that the kinetic energy of the electron (say in the electron beam) is equal to the energy acuired by being accelerated through the voltage V (20,000V). Thus:

a) ½ mv^2 = qV = eV

where q = e is the electronic charge

Then the velocity of a given electron in the beam will be:

v = [2 eV/m]^1/2

but note here the quantity (e/m) is just the electron charge to mass ratio which can be obtained from a table of fundamental constants as: (e/m) = 1.8 x 10^11 C/kg

therefore, v = [ 2 (1.8 x 10^11 C/kg) (20,000 V)]^1/2 = 8.5 x 10^7 m/s

To get the deflection of the beam (y) we need to find the time taken to reach the screen first.

t = x/ v = 0.4m / (8.5 x 10^-7 m/s) = 4.7 x 10^-9 s

Then the deflection can be obtained using the basic kinematic form:

y = ½ at^2

where a is the acceleration but incepted by the effect of the Earth's magnetic field B.

For a given magnetic field, the force exerted is F(B) = qvB = evB

In this case:

F(B) = (1.6 x 10^-19 C) (8.5 x 10^7 m/s) (5 x 10^-5 T) = 6.8 x 10^-16 N

This is the force applied by the Earth's magnetic field to the electrons in the beam

By Newton's 2nd law: F = ma

ma = e vB so the acceleration a = e vB/ m = (e/m) v B

where (e/m) is the charge to mass ratio already obtained earlier, so:

a = (1.8 x 10^11 C/kg) ((8.5 x 10^7 m/s) (5 x 10^-5 T) = 7.6 x 10^14 m/s^2

Then, the displacement is found from:

y = ½ at^2 = y = ½ ( 7.6 x 10^14 m/s^2) ( 4.7 x 10^-9 s) = 8.4 x 10^-3 m

or 8.4 mm

(9) We have for force balance:

F = mg = u(o) I1 (I2) L/ 2 pi r

Where u(o) is the magnetic permeability = 4 pi x 10^-7 H/m

Since the same current runs through parallel wires on the balance, I1 = I2 = I

so: I1(I2) = I^2

and we can solve for the current needed to balance the force (mg) = 10N

I = [ 2 pi r (mg)/ u(o) L]^1/2

I = [ 2 pi (0.1m) (10N)/ (4 pi x 10^-7 H/m)(1m)]^1/2

and I = 2.2 x 10^3 A

10) The new half life curve, for T( ½ ) = 250,000 years is shown

After three million years, how much of the original sample will be left?

After one half life (250 ky) it is ½ gram so this is reduced by halves over successive half lives

Then we have:

After 2T( ½ ): 1/4 g

After 3T( ½ ): 1/8 g

After 4T( ½ ): 1/16 g (One million years)

After 5T( ½ ): 1/32 g

After 6T( ½ ): 1/64 g

After 7T( ½ ): 1/128 g

After 8T( ½ ): 1/256 g (Two million years is eight half lives)

After 9T( ½ ): 1/512 g

After 10T( ½ ): 1/1024 g

After 11T( ½ ): 1/2048 g

After 12T( ½ ): 1/ 4096 g

so the amount left after 3 million years = 1/ 4096 g

= 2.4 x 10^-4 g or 0.24 mg (millgram)

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