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Physics: Post your doubts here!

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Laila39

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A laser emits light of wavelength 600 nm.
What is the distance, expressed as a number of wavelengths, travelled by the light in one second?
A - 5 × 10^8
B 5 × 10^11
C 5 × 10^14
D 5 × 10^17
How do I solve this? What formula is being applied. :/
 
mehria

mehria

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Student12 said:
Do some of them at least :( or by tonight.. :/ i know you have to prepare.
Click to expand...
Question 3, 4 , 5 , 8 , 11 , 14 , 15 , 19 , 24 , 33 , 35 , 37 , 38 PLEASEEEE!http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_12.pdf

Q3:- as Cd has no units the we won't mention it in our sol.
so ρ (V^n) A = F
V^n = F/ρA
now we will just put the base units of them...
(m/s)^n = (kgm/s^2) / (kg/m^3 x m^2)
(m/s)^n = ( m^2 / s^2 )
(m/s)^n = (m/s)^2
so n=2

Q4:-
Screenshot (939).png

Q5:- for ths question m gettng B :/ but still i'll try it once again n then tell u

Q8:-

Q11:- here the first we have to do is to change the cm^2 of A into m^2
so for that we will divide 1 by 10000 = 1 x 10^-4
then we will put all the values in the formula:- P= F/A
over here as we r given with the amount of a-particles colliding per second then by multiplying ths value with speed will give us the acceleration...
so P= ((6.6 x 10^-27) x (1.5 x 10^7) x (5.0 × 10^4)) / (1x10^-4)
= 4.95 x 10^ -11

Q14:- in ths question we will simply find the resultant force of both at X n Y and the subtract them frm eachother
the resultant force in X is :- 2(Tcosθ) = 2 ( 100 cos65) = 84.6
the resultant foce in Y:- 2(Tcosθ) = 2( 120 cos55) = 137.6
so the increase will be :- 137.6 - 84.6 = 53

Q15:-

Q19:- we r given with volume and density n we have to find the mass first..
so m=1000 x 340 = 340000 kg
time= 60 s ( 1 minute)
g= 9.81
h= 30m
put ths in the equation :- P= (mgh) /t to get the input power
our input power = 1667700 W

as efficiency= (output power / input power) x 100
then output power= 1500930 W = 1.5 MW

Q24:- F = ke
so e= F/k
e= 25/150
e=0.167
so original length= 0.55 - 0.167 = 0.383 m

Q33:- P=I^2R ( m using ths formula cuz current will be the same acroos both resistors as they r in series where as V will be dffrnt)
so P in internal resistor = I^2 R
in external resistor = (I^2) 2R
so their ration will be 2

Q35:-

Q37:- In ths question we have to check for the resistance... we know that when resistance is less then voltage will be also less
so there will be lesser voltage in resistor 2 Ω and will be greater in resistor 4 Ω
n as the voltage from X to Y is frm positive to negative so it will be decreasing passing each resistor

Q38:- ok so first we will find the total V across resistors R, 2R, 3R
V in R= 2 V
total resistance= 6R
so
Vr = (2/6) V (Vr is the V across resistor 2R)
so V=6
as the V remains the same in parallel circuits so V across 4R will be also 6V
n emf = 6V
 
D

Dr.MMM

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sagar65265 said:
First Question)

Okay, so what we need to find out is the percentage uncertainty in the value of the volume of the sphere with a radius measurement of r = 5.00 mm ± 0.01 mm.

The formula for volume of a sphere as a function of radius is 4πr³/3. Therefore, we are multiplying r into r into r to get r³, and this is the only value in the equation that has any uncertainty - there is no uncertainty in 4/3 and there is no uncertainty in π, so r is the only quantity we need to concern ourselves with.

The rule for uncertainties when any number of quantities are multiplied goes as follows:
"When one multiplies or divides several measurements together, one can often determine the fractional (or percentage) uncertainty in the final result simply by adding the uncertainties in the several quantities."

Let's do this by example. We are multiplying r by itself three times (to get r³) and so to get the percentage uncertainty in the final result, we resort to "adding the uncertainties in he several quantities". In other words, we add the (percentage uncertainty in r) to the (percentage uncertainty in r) to the
(percentage uncertainty in r).

Basically we multiply the percentage uncertainty in r by 3.
The percentage uncertainty in r = (0.01 mm/5.00 mm) * 100 = 0.2 %.
Multiplying this by 3, we get the uncertainty in the volume of the sphere with this radius to be 0.2 * 3 = 0.6% = C.

(more information here: http://spiff.rit.edu/classes/phys273/uncert/uncert.html)

Second Question)

Let's do the math behind this - suppose the distance traveled with both speeds is denoted by "s".
Suppose the time taken at 600 kmph = t(1).

Then, since Speed = Distance/Time, we can write

600 = s/t(1)
So that
t(1) = s/600

Similarly, let's do the same for the other speed. The distance traveled is still "s", the speed = 400 kmph, and the time taken to traverse this distance = t(2). So:

400 = s/t(2)
So that
t(2) = s/400

The average speed is the total distance traveled dividing by the time taken (the average velocity is different - it is equal to total displacement divided by time taken).
= (Total Distance traveled)/(Time taken to cover that distance).

Since the plane travels a distance "s" in one direction and returns the same distance "s", the total distance traveled = s+s = 2s.

The time taken = t(1) + t(2) = s/400 + s/600 = 3s/1200 + 2s/1200 = 5s/1200 = s/240

Therefore, the average speed = 2s/(s/240) = 480 kmph = C.

Third Question - Q22)

https://www.xtremepapers.com/commun...st-your-doubts-here.9860/page-554#post-826497

Fourth Question - Q23)

From the earlier question I think you might see a pattern - the equivalent spring constant of springs in series (end-to-end attachment) is found in a similar manner to the way the equivalent resistance of resistors connected in parallel is found:

1/k(equivalent) = 1/k(1) + 1/k(2) + 1/k(3)..........+ 1/k(n) (when n springs of spring constant k(1), k(2), k(3) until k(n) are connected in series)

Similarly, the equivalent spring constant of springs connected in parallel (connected side to side at the same level) is found in a similar manner to the way the equivalent resistance of resistors connected is series is found:

k(equivalent) = k(1) + k(2) + k(3).............+ k(n) (when n springs of spring constant k(1), k(2), k(3) until k(n) are connected in parallel)

So, what we can do here is find the equivalent spring constant for each option, and see which one turns out to have the largest extension.

Let's take A first. There are two coils, which implies that there are two springs in that combination, connected in series. Also, both those springs have the same spring constant, let's say "k". Therefore, we can add their spring constants like so:

1/k(equivalent) = 1/k + 1/k = 2/k
Therefore, since 1/k(equivalent) = 2/k, k(equivalent) = k/2.
So the spring constant for the setup shown in option A is equal to k/2.

Onto option B. There are three coils, so we can assume that there are three springs connected in series, and each one of them has the same spring constant, "k" - the same spring constant as the one used for the previous option. So, their equivalent spring constant is given by

1/k(equivalent) = 1/k + 1/k + 1/k = 3/k
Since 1/k(equivalent) = 3/k, k(equivalent) = k/3.
So the spring constant for the setup shown in option B is equal to k/3.

Option C: We have two springs in parallel. Finally, an easier calculation!:D
Either ways, each one those springs has the spring constant "k", again the same as before. Since they are attached in parallel, we can find their equivalent spring constant as follows:

k(equivalent) = k + k = 2k.

Option D: Happy days, another simple calculation!:)
Okay, so there are three springs in parallel, and all of them have the spring constant k. So, their equivalent spring constant is given by

k(equivalent) = k + k + k = 3k.

Alright. Now for the extensions.
Each of those springs follows the formula |F| = kx, or (Magnitude of Stretching Force) = (Spring Constant) * (Extension). So, we can rewrite the equation as
x = |F|/k. Let's calculate these ratios for each of the options.

Option A: x = (2 Newtons)/(k/2) = (4/k) meters.
Option B: x = (1 Newton)/(k/3) = (3/k) meters.
Option C: x = (6 Newtons)/(2k) = (6/2k) = (3/k) meters.
Option D: x = (8 Newtons)/(3k) = (8/3k) = (2.667/k) meters.

Clearly, out of all these, the largest extension comes about for Option A, since the rest all have the same denominator but a smaller numerator. So, A.

Hope this helped!
Good Luck for all your exams!
Click to expand...
Thanks A LOT.
It really helped :)
Physics ques.png
ans is B. How?
 
danial 234

danial 234

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Laila39 said:
A laser emits light of wavelength 600 nm.
What is the distance, expressed as a number of wavelengths, travelled by the light in one second?
A - 5 × 10^8
B 5 × 10^11
C 5 × 10^14
D 5 × 10^17
How do I solve this? What formula is being applied. :/
Click to expand...
in one second light will travel 3 × 10^8 metres
1 wavelength= 600 nm
x wavelength= 3 × 10^8 metres
x= 3 × 10^8 metres/600 nm
 
S

Shaoli Hassan

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Can someone please help me out with October/November 2011/variant-12, Question Number 10? The correct option is A, but I do not understand why.
 
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