I dont understand no3. You said the angle of incidence is 37 degrees(the critical angle) but to answer the question on why red light doesn't change direction, you said "angle of incidence is 0"
Could you explain question no.6(iii). If the incident angle of blue light is less than the critical angle, then how does it undergo total internal reflection.
The critical angle for blue light has a smaller magnitude than 37 degrees. When blue light is incident at 37 degrees, the angle of incidence is greater than critical angle for blue light, hence total internal reflection occurs.
@@ak-ec7yp Basically you just have to find the Tension using the formula CWmoment=ACWmoment. We use the feet of the climber as the pivot. The clockwise moment (downward) is, moment=Fd, which is 0.90m x 620N(her weight). To get the ACWmoment, you must find the Tension that is perpendicular to the body (instead of 45degree slope) ,draw out a triangle I\ T(perpendicular)I \T(slope) I___\ 1.2m T(perpendicular)=Tsin60 Now you can find T CWmoment=ACWmoment Fd=Fd 0.90(620)=Tsin60(1.20 T=540N Btw im taking the exam too (in Hanchiang) Hope that helps!
that's how you got the equation moment=Fd ^(force perpendicular from pivot) ________________| /\ the tension given in the question was a slope can it cant be used in the equation moment=F x perpendicular distance from pivot angle is given so you can use it to find the F(perpendicular) in order to calculate the ACWmoment
In this question you can find the half-life by the time taken for count rate to reduce by half. You begin by subtracting the background count rate (which is due to background sources and not from the radioactive sample itself) from any value from graph, find its half then add back the background count. Lastly find the time for the change from your first chosen count rate value and your final count rate.
I dont understand no3. You said the angle of incidence is 37 degrees(the critical angle) but to answer the question on why red light doesn't change direction, you said "angle of incidence is 0"
Hey can you explain 9 A) a bit
for question 6b iii) how is the critical angle for red light 37 degrees if there is still light being reflected
When angle of incidence is equal to the critical angle, some of the light is reflected and some of the light is refracted.
Could you explain question no.6(iii). If the incident angle of blue light is less than the critical angle, then how does it undergo total internal reflection.
The critical angle for blue light has a smaller magnitude than 37 degrees. When blue light is incident at 37 degrees, the angle of incidence is greater than critical angle for blue light, hence total internal reflection occurs.
Thank you@@examhelpweb2090
Can you pls explain 3 c ii again
I find it difficult to understand
Can you explain it simpler again
I’ve got my igcse in 3 days
Did you find the explanation for this ? I have my exam too if u have can you please share it
@@ak-ec7yp Basically you just have to find the Tension using the formula CWmoment=ACWmoment.
We use the feet of the climber as the pivot. The clockwise moment (downward) is, moment=Fd, which is 0.90m x 620N(her weight). To get the ACWmoment, you must find the Tension that is perpendicular to the body (instead of 45degree slope) ,draw out a triangle I\
T(perpendicular)I \T(slope)
I___\
1.2m
T(perpendicular)=Tsin60
Now you can find T
CWmoment=ACWmoment
Fd=Fd
0.90(620)=Tsin60(1.20
T=540N
Btw im taking the exam too (in Hanchiang)
Hope that helps!
@@kun9444 what do you mean by the tension perpendicular to the body, thats my confusion
that's how you got the equation moment=Fd
^(force perpendicular from pivot)
________________|
/\
the tension given in the question was a slope can it cant be used in the equation moment=F x perpendicular distance from pivot
angle is given so you can use it to find the F(perpendicular) in order to calculate the ACWmoment
did you pass bro?
nice
i really support
in number 9 y did u use 390
I didn't have to. You can pick any value from them graph and find the time taken for it to reduce by half the amount.
u dont answer or reply to ur comments
i did not get no 9a)iii properly
In this question you can find the half-life by the time taken for count rate to reduce by half. You begin by subtracting the background count rate (which is due to background sources and not from the radioactive sample itself) from any value from graph, find its half then add back the background count. Lastly find the time for the change from your first chosen count rate value and your final count rate.
you titled the video wrong ,its IGCSE not A level
It's correct
@@mehboob-ulhassan8882 they edited it