^{7}m/s enters midway between two horizontal plates in a direction parallel to the plates which are 5 cm long and 2 cm apart, and have a potential difference V between them. Find V, if the beam is deflected so that it just grazes the bottom plate. (Take the electron charge to mass ratio: e/ m

_{e}= 1.8 x 10

^{11}C/kg).

Solution:

The
key piece of information (and key to the solution!) is that the beam

E = V/d = V/ (0.02m)

The downward acceleration, fromNewton 's
2nd law, is:

*just grazes the bottom plate*. Since the total y -span = 2 cm, this means the effective beam deflection is one half this, or dy = 1 cm = 0.01m. Meanwhile, the electric field is:E = V/d = V/ (0.02m)

The downward acceleration, from

a

The vertical displacement in terms of this acceleration is:

y = ½ {a

The time, t, applicable is:

t = x/v = (0.005m)/ 10

and we know:

y = ½ (V/d)(e/ m

Substituting:

0.01m = 0.5(V/0.02m) [1.8 x 10

for which V can be found: V = 89 V

_{y}= eE/m_{e}The vertical displacement in terms of this acceleration is:

y = ½ {a

_{y}t^{2}) = ½ {eE/ m_{e}} t^{2}The time, t, applicable is:

t = x/v = (0.005m)/ 10

^{7}m/s = 5 x 10^{-9}sand we know:

y = ½ (V/d)(e/ m

_{e })t^{2}Substituting:

0.01m = 0.5(V/0.02m) [1.8 x 10

^{11}C/kg] (5 x 10^{-9}s)^{2}for which V can be found: V = 89 V

2) A
heated filament emits electrons which are accelerated to the anode by a p.d. of
500 V. Find the kinetic energy and velocity of the electron as it strikes the
anode.

__Solution__:

We
have V = 500 V

But
the electric field E is defined: E = 500V /d

Where
d is the distance between plates

The
acceleration will be: a = eE / m

_{e}= e (500V /d )/ m_{e}
Where
(e/ m

_{e}) is the charge to mass ratio.
Then v

_{y}= a_{y}t = (eE/ m_{e}) d/v_{y}= [e (500V /d )/ m_{e}] d/ v_{y}= e (500V )/ m_{e}] v_{y}
And:

(v

_{y})^{2}= e (500V )/ m_{e}
Where the charge to mass ratio is: e/ m

_{e}= 1.8 x 10^{11}C/kg
So:
(v

_{y})^{2}= (1.8 x 10^{11}C/kg) (500V ) = 9 x 10^{14 }m^{2}/ s^{2}^{ }

And: v

_{y}= 3 x 10^{7}^{ }m/ s
The
KE is: ½
m

_{e }(v_{y})^{2}= ½ (9.1 x 10^{-31}kg) ( 9 x 10^{14 }m^{2}/ s^{2 })
= 4.0 x 10

^{-16}J^{}
3) Given
that x' = 1/a (x - vt) and t' = 1/a (t - vx/c

^{2}), derive similar equations for x and t in terms of x' and t'. (Let: 1/a = (1 - v^{2}/c^{2})^{½})
If
we now substitute x' = a(x - vt) and equation for t’ into the right hand side of:

r’

^{2}= x’^{2}+ y’^{2}+ z’^{2}= c^{2}t’^{2}^{}
what
do we get?

__Solution:__

Given
x' = 1/a (x - vt) and t' = 1/a (t - vx/c

^{2}),Then: x' = x/a - vt/a and t' = t/a - vx/ac

^{2}

and: x' + vt/a = x/a and t' + vx/ ac

^{2 }= t/a

so:

x = a(x' + vt/a) and t' = a(t' + vx/ ac

^{2 })

finally: x = a(x' + vt) and t = a(t' + vx/c

^{2})

b)
On substituting x'
= a(x - vt) and equation for t’ into the
right hand side of:

r’

^{2}= x’^{2}+ y’^{2}+ z’^{2}= c^{2}t’^{2}
We
get:

x

^{2}+ y^{2}+ z^{2}- c^{2}t^{2}= a^{2}(x - vt)^{2}+ y^{2}+ z^{2}
4) Let
a quadratic model for central meridian transit of a sunspot group – indicating
days past the central meridian (where the closer the value is to 0, the greater
the incidence of likely flaring) be:

F(x)
= 0.1x

^{2}+ 8x + 12
Is
this complex group more or less likely to see powerful flares than the one
given in the example?

__Soln__.

If it is presumed to incept more powerful flares it should be fewer days from the central
meridian.

Solving
using

*the quadratic formula*:
[-b

__+__{b^{2 }- 4ac}^{1/2}]/ 2a
Where a = 0.1, b = 8 and c = 12

x1
= -1.529 and x2= - 78.471

Choosing
the lesser of the two and comparing it the lesser value in the text we find the
latter has x1 = -1.396 so it is more likely to produce powerful flares since the
group is more proximate to the central meridian.

5)
An athlete
standing close to the edge on the top of a 160 ft. high building throws a
baseball vertically upward. The quadratic function:

s(t)
= -16 t

^{2}+ 64t + 160
models
the ball’s height above the ground, with s(t) in feet, t seconds after it’s
thrown:

a)After how many seconds does the ball reach its
maximum height? What is the maximum height?

b)How many seconds does it take until the ball
finally hits the ground? (Round to the nearest tenth of a second)

__Solution__: We plot a

**of s(t) vs. t with t on**

*graph**the horizontal axis*. We find on doing so that we obtain a parabola with a maximum at t = 2 s, and s(t) = 264’

Hence,
the answer to (a) is t = 2 seconds and the maximum ht. = 264 ft.

To
answer (b) we examine the graph to see when the motion ends (e.g. where
parabola terminates with s(t) = 0) and
find it to be at t = 5.7 secs. This can also be found from the quadratic formula, setting s(t) = 0 and solving for t.

6) Another
form of Newton ’s
law of gravitation which incorporates Kepler’s third law is:

P

^{2}= 4(p)^{2}a^{3}/[G(M + M_{p})].
where
P is the planet's period, a the distance (i.e. semi-major axis, we want to
solve for) and G the gravitational constant, M the mass of the Sun, and M

_{p}the planet's mass.
Write
an equation for the planet's mass, i.e. which might be used to find it if all
the parameters are known, and the units.

__Solution__:

Cross
multiply to get:

P

^{2 }[G(M + M_{p})] = 4(p)^{2}a^{3}
Then
divide both sides by P

^{2}:
[G(M
+ M

_{p})] = 4(p)^{2}a^{3 }/ P^{2}
Divide
through by G:

M
+ M

_{p}= 4(p)^{2}a^{3 }/ G P^{2}
Solve
for M

_{p}:
M

_{p}= 4(p)^{2}a^{3 }/ G P^{2 }- M*Partial Fraction Decomposition Problem:*

Add
the following algebraic expressions and arrive at such a single rational
expression:

3/
(x – 4) + 2/ (x + 2)

While
this is not an application per se, it IS frequently one of the algebraic
manipulation tasks needed in calculus.

Then the
common denominator is :

(x – 4) (x + 2) = x

(x – 4) (x + 2) = x

^{2}- 2x -8
So
we can write:

3
(x + 2) + 2 (x – 4)/ (x

^{2}- 2x -8)
Or:

[3x
+ 6 + 2x – 8]/ (x

^{2}- 2x -8) = 5x -2/ (x^{2}- 2x -8)
## No comments:

Post a Comment