A car moves in a bend in a road of angle tan-1 5/12 and radius 63.2 . If the coefficient of friction is 0.4 find
1. The maximum velocity that will keep the car in motion and to keep it from skidding
2. The minimum velocity

Danielgabby:
A car moves in a bend in a road of angle tan-1 5/12 and radius 63.2 . If the coefficient of friction is 0.4 find
1. The maximum velocity that will keep the car in motion and to keep it from skidding
2. The minimum velocity

Cc oxytocin

What is the critical angle of light passing from a material of index of refraction 1.54 to a material of index of refraction n=1.33
Oxytocin Geofavor mathefaro

Danielgabby:
A car moves in a bend in a road of angle tan-1 5/12 and radius 63.2 . If the coefficient of friction is 0.4 find
1. The maximum velocity that will keep the car in motion and to keep it from skidding
2. The minimum velocity

Now. I don't really know the difference between the maximum and minimum velocities.
But here's what I can solve
Modified: what are we to do with the coefficient of friction? cc Geofavor

.

ShyNeptune:
What is the critical angle of light passing from a material of index of refraction 1.54 to a material of index of refraction n=1.33
Oxytocin Geofavor mathefaro

Refractive index of medium of incidence ray=1.54

1.54=1/SinC


SinC=1/1.54

SinC =0.6494

C=Sin^-1(0.6494)

C=40.49~40.5°

Oxytocin:

Refractive index of medium of incidence ray=1.54

1.54=1/SinC


SinC=1/1.54

SinC =0.6494

C=Sin^-1(0.6494)

C=40.49~40.5°

thanks but .... but the answer here is 59.7 degree and i don't know how it was gotten

remember two Refractive indices were given.

ShyNeptune:

thanks but .... but the answer here is 59.7 degree and i don't know how it was gotten

remember two Refractive indices were given.

Hmm... Owkk
Will check it out again

ShyNeptune:

thanks but .... but the answer here is 59.7 degree and i don't know how it was gotten
remember two Refractive indices were given.

Did some search....
This came up..


ni/nr =1/SinC
SinC =nr/ni
C= Sin^-1(nr/ni)

BTW.... From a reliable source ***

Θcrit= sine-1 (nr/ni)

The critical angle can be calculated by taking the inverse-sine of the ratio of the indices of refraction. The ratio of nr/ni is a value less than 1.0


So..

C=Sin^-1(1.33/1.54)
C=Sin^-1(0.864)
C=59.7°

A satellite moves in a circular orbit of radius 4R round the earth. The acceleration of the satellite in terms of g is?


Thankyoujesus, mathefaro, shyneptune, oxytocin

Oxytocin:

Did some search....
This came up..


ni/nr =1/SinC
SinC =nr/ni
C= Sin^-1(nr/ni)

BTW.... From a reliable source ***

Θcrit= sine-1 (nr/ni)

The critical angle can be calculated by taking the inverse-sine of the ratio of the indices of refraction. The ratio of nr/ni is a value less than 1.0


So..

C=Sin^-1(1.33/1.54)
C=Sin^-1(0.864)
C=59.7°

thanks a million

Geofavor:
A satellite moves in a circular orbit of radius 4R round the earth. The acceleration of the satellite in terms of g is?


Thankyoujesus, mathefaro, shyneptune, oxytocin

A metal wire of mass 1 gram and length 50cm is under a tension of 80N. Calulate the fundamental frequency


abeg, my ogas, help me solve this things asap. No time!

Thankyoujesus, mathefaro, shyneptune, oxytocin etc.

Geofavor:


A metal wire of mass 1 gram and length 50cm is under a tension of 80N. Calulate the fundamental frequency


abeg, my ogas, help me solve this thing.

11.25.

??

ShyNeptune:
11.25.
??

Lol. Was it a guess work you did? That's not the answer. It's far from it
Btw, what's your exam date?

Geofavor:


A metal wire of mass 1 gram and length 50cm is under a tension of 80N. Calulate the fundamental frequency


abeg, my ogas, help me solve this things asap. No time!

Thankyoujesus, mathefaro, shyneptune, oxytocin etc.

I think the formula to be used is: f_o= 1/2L||T/M

|| Means square-root

Orezy5:
I think the formula to be used is: f_o= 1/2L||T/M
|| Means square-root

yes it is. Kindly solve it step by step. Thanks, bro

Geofavor:

Lol. Was it a guess work you did?
That's not the answer. It's far from it

Btw, what's your exam date?

I didn't guess
I used the formula...
One over two pie times square root of gravity over massDensity


My exam? 3rd March

Geofavor:

yes it is. Kindly solve it step by step. Thanks, bro

f_o= 1/2L||T/M

f_o=?

L= 0.5m

T=80N,

M= mass per unit length(Kgm^-1)= 0.001/0.5=0.002kg/m

f_o= 1/2*0.5||80/0.002

f_o=1||40000

f_o= 200Hz

|| means square-root

I just saw that tension was given....
I was busy calculating frequency of a plucked string

Orezy5:

f_o= 1/2L||T/M
f_o=?
L= 0.5m
T=80N,
M= mass per unit length(Kgm^-1)= 0.001/0.5=0.002kg/m
f_o= 1/2*0.5||80/0.002
f_o=1||40000
f_o= 200Hz
|| means square-root

Boss respect

ShyNeptune:
I just saw that tension was given....
I was busy calculating frequency of a plucked string


Boss respect

Boss ke?

Orezy5:

Boss ke?

ShyNeptune:

I didn't guess
I used the formula...
One over two pie times square root of gravity over massDensity


My exam? 3rd March

Wow
March 3rd same date as mine, where is your center??

Geofavor:
A satellite moves in a circular orbit of radius 4R round the earth. The acceleration of the satellite in terms of g is?


Thankyoujesus, mathefaro, shyneptune, oxytocin

The velocity of a satellite moving in a circular orbit round the earth is giving by

V=√(gr)

Since r = 4R

So V=√(4Rg)

V² = 4Rg

g =V²/4R

ShyNeptune:

thanks a million

You welcome bro

Oxytocin:

Wow
March 3rd same date as mine, where is your center??

Cool...My Date Mate
My exam center? Somewhere in Awka....

ShyNeptune:

Cool...My Date Mate
My exam center? Somewhere in Awka....

Lolz... Awka??
I'm in Lagos o

The difference no be for here...

Oxytocin:
Lolz... Awka?? I'm in Lagos o
The difference no be for here...

Lolz

a beg i day live for orile n my exam centre na aguda at shoms intern school for odutola abeg he I fit locate DAT Syd

Just reprinted today and am writing on the 1st of march. GOD AM SO DONE. Pray For Me Friends

iceberryose:
Just reprinted today and am writing on the 1st of march. GOD AM SO DONE. Pray For Me Friends

Go and come back with nothing but good news

Orezy5:
f_o= 1/2L||T/M
f_o=?
L= 0.5m
T=80N,
M= mass per unit length(Kgm^-1)= 0.001/0.5=0.002kg/m
f_o= 1/2*0.5||80/0.002
f_o=1||40000
f_o= 200Hz
|| means square-root

thanks

Geofavor:
thanks

Welcome