^{6}m/s is much less than c, the speed of light. Thus:

dv/dt = a and upon integration:

v - v(o) = at

and dx/dt = v

Integrating a second time one obtains:

x - x(o) = v(o)t + ½ at

^{2}

Now: a = -1.2 x 10

^{14}m/s/s

v(o) = 6 x 10

^{6 }m/s and the final velocity is: v(f) = 0

So: v(f) - v(o) = at = 0 - 6 x 10

^{6}m/s = (-1.2 x 10

^{14}m/s/s) t

whence: t = [0 - 6 x 10

^{6}m/s ] / (-1.2 x 10

^{14}m/s/s)

t = 5 x 10

^{-8}sec

Now: x(f) - x(o) = v(o)t + ½ at

^{2}

= (6 x 10

^{6}m/s)(5 x 10

^{-8}sec) - ½((-1.2 x 10

^{14}m/s/s)(5 x 10

^{-8}sec)^2

x(f) - x(o) = 0.15 m =

__15 cm__

17) By Newton's law of universal gravitation, the force between the two masses is:

GMm /r

^{2}

From the integration of the above the potential energy can be obtained as: -(GMm/R) and the kinetic energy as KE = GMm/2R.

let d = 2R then the total KE available is:

GMm/d

Now: Momentum must also be conserved so (assuming conservation of linear momentum along the line of action):

mv(m) = M v(M)

But: KE = ½ [mv(m)

^{2}+ M v(M)

^{2}] = GMm/d

Using algebra:

v(M)

^{2}[1 + M/m] = 2GM/d and

v(m)

^{2}+ v(M)

^{2}+ (m/M)v(m)

^{2}+ (M/m)v(M)

^{2}= 2G/d{M + m}

v(m)

^{2}+ v(M)

^{2}= 2G/d (M + m)

Therefore: v

_{R }=

__[2G/d (M + m)]__

^{`1/2}

18) We define current as the rate at which charge flows, thus:

I = dQ/ dt so dQ = I dt, or Q = It

since I = 5 A (amps) = 5 coulombs/sec

then for 4 minutes (= 4 x 60 sec = 240 secs):

Q = 5 coulombs/sec (240 sec) =

__1200 coulombs__

19) The effective value of a sinusoidal emf is its root mean square value, defined:

v(eff)

^{2}= v(rms)

^{2}= ∫ [v(max)

^{2}sin wt dt] T = ½ v(max)

^{2}

(integrated from 0 to t)

so v(eff) = [½ v(max)

^{2}]

^{1/2}= 1/ [2]

^{1/2}(v(max)) or

v(eff) =

**Ö**2 (v(max)) / 2

20) Potential difference is the work needed to move a test charge (q) from point A to B, or:

V = W/q

In this case, W = 2 J and q = 20 C, so:

V = 2J/ 20 C = 0.1 Volt

21) For a parallel plate capacitor,

C = k k(o) A/ d

where k(o) = 1 for air, and k is the relative dielectric constant: or

k = k(mica) k(o) = 6 k(o) = 6

Then C(mica) = 6 (15 m mF) = 90 m m F

(A/d is the same in each case)

22) The charge on the electron, q = 1.602 x 10

^{-19}C

Again, from Problem #18, the current is charge moved per unit time:

I = dQ/ dt

In this case, since the angular frequency is the key information given, and it is related to time via:

w = 1/T

then:

I = q w = (1.602 x 10

^{-19}C) (6.6 x 10

^{15}/sec)

I = 1.05 x 10

^{-3}amp, or I = 0.00105 A

23) For a plane mirror the applicable law of geometrical optics is that the image appears to be as far behind the mirror as the object is in front. Therefore, one's eyes must be focused to a distance of 40 cm.

24) We apply the

*lens makers' formula*here, see e.g. http://brane-space.blogspot.com/2011/06/introduction-to-basic-physics-thin.html:

1/ f = (n - 1) [1/r1 - 1/r2]

where f is the focal length of the convex lens (to be found)

n = 1.65 (the refractive index) and r1 = 40 cm

then r2 = -40 cm (recall for convex lenses, r2 = -r1)

Then:

1/f = (1.65 - 1) [1/40 cm

**-**(1/ -40cm)]

1/f = 0.65/ 20 cm

So: f = 20 cm/ 0.65 = 31 cm

25) In this nuclear reaction, the charge and nucleons

*must be conserved*, meaning that the upper index total on the right side must be equal the upper index total on the left, plus the lower number on the right must be equal to the sum of lower numbers on the left. The only one which fits this is

__(D__)

8 O 17.

26) Here, we use the relativistic form for the kinetic energy:

K= Mc

^{2 }

**-**

^{ }M(o)c

^{2}

We are given the

*rest mass*is 106 MeV so M(o)c

^{2}= 106 MeV

We also have: K = 4 MeV, then:

Mc

^{2}= K + M(o)c

^{2}= 4 MeV + 106 MeV = 110 MeV

We also know:

M = M(o)/ [1 - (v/c)

^{2})]

^{1/2}

But K = M(o)c

^{2}[1/ (1 - (v/c)

^{2})]

^{1/2}) - 1]

Therefore (with some algebraic working):

(v/c) = 0.28

or

v = 0.28c =

__84,000 km/sec__

27) The system depicted in the graphic below is the one used for solution here.

The Lagrangian is the difference between kinetic and potential energies:

L = T - V

And from the system:

T = ½ m1v1

^{2}+ ½m2v2

^{2}= ½ (m1 + m2)x'

^{2}

V = (m1 x - m2 x) g

So:

L = ½ (m1 + m2)x'

^{2}- (m1 x - m2 x)g

= ½ (m1 + m2)x'

^{2}+ (m2 x - m1 x)g

28) Applying force balance:

m2 g - T = m2 a

and:

m1 g - T = -m1a

(where

**T is the tension in the string**.)

so: Using algebra to solve the simultaneous equation:

m1 m2 g - m1 T = m1 m2 a

m1m2 g - m2T = -m1m2 a

----------------------------

2 m1m2g - (m1 + m2) T = 0

and

__T = (2 m1m2 g) / (m1 + m2)__

29) The efficiency of a Carnot cycle engine is:

e = 1 - Q(R) / Q(W)

where Q(R) denotes the heat returned to a reservoir at T2, and Q(W) is the heat withdrawn from a reservoir at T1.

Then:

Q(R)/ Q(W) = T2 / T1

and

e = 1 - Q(R) / Q(W) = 1 - (T2/T1)

30) From the quantum principle for the total angular momentum:

L = L1 + L2........[L1 - L2]

Thus:

L = 5, 4, 3, 2, 1

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