[color=#006600][/color] chemistry update come up ........... Physics update comes up later in d day

mathefaro:
It's a mathematics question
If the bearing B from A is 030, then the bearing of A from B is 210^o or S30^oW.

thank u @mathefaro
ok...but I found d question via physics textbook

Destilo:


thank u @mathefaro
ok...but I found d question via physics textbook

like I said in the other thread, bearing is bearing irrespective of where you've seen it

mathefaro:
like I said in the other thread, bearing is bearing irrespective of where you've seen it

thank u

Benjaminbest:
[color=#006600][/color] chemistry update come up ........... Physics update comes up later in d day

I think its time for physics update...or isn't it?

1. A rough wooden block on a rough plank is drawn along the plane of the plank when inclined at 60 degree to the horizontal. Determine the coefficient of friction between two surfaces.

2. If a force of 50N moves a load of 30kg up an inclined plane. Determine the magnitude of the force if the plane is inclined at an angle of 30 degree.


Thank you

Make una no blame me if i no update glo network jst Bleep me up am using somebodys fone na

guys before i go update....make una tell me d differences between moving coil galvanometer and hot iron ammeter. Guys before u una resonate with my question make una wave for mind so dat u guys will b frequent with maximum velocity. Jst before i accelerate to d next level make my weight no become scalar

Benjaminbest:
guys before i go update....make una tell me d differences between moving coil galvanometer and hot iron ammeter. Guys before u una resonate with my question make una wave for mind so dat u guys will b frequent with maximum velocity. Jst before i accelerate to d next level make my weight no become scalar

Moving coil Galvanometer is an instrument that is used for the DETECTION of electric currents through electric circuit..

and

Hot Iron Ammeter is an instrument used to MEASURE the flow of electric current in a circuit.

Benjaminbest:
guys before i go update....make una tell me d differences between moving coil galvanometer and hot iron ammeter. Guys before u una resonate with my question make una wave for mind so dat u guys will b frequent with maximum velocity. Jst before i accelerate to d next level make my weight no become scalar

moving coil galvanometer responds only to direct currents(DC), it is used for detecting very small amounts of current the current to be measured is sent through a coil suspended or pivoted in a fixed magnetic field, and the current is determined by measuring the resulting motion of the coil.

while hot wire ammeter is an ammeter which measures alternating or direct current by sending it through a fine wire, causing the wire to heat and to expand or sag, deflecting a pointer. Also known as thermal ammeter

pls kindly help me in this...@mathefaro n other maths expert in d house...

Express 110011.1001 base 2 to
1. Octal
2. Hexadecimal
3. Base 4


I got the whole number side...but find it hard to get the fractional parts...after converting it to base 10, can't convert d fractional parts to the bases above..

Thank you very much

gudmuni my people...happy sunday. Let me start with apology am greatly sori for disappointing u guys again. I have loss many fans due to lack of serious. Dis thread will now b functioning everyday. Today is our cultural harvest u all are invited. Hope iam forgiven

Destilo:
pls kindly help me in this...@mathefaro n other maths expert in d house...

Express 110011.1001 base 2 to
1. Octal
2. Hexadecimal
3. Base 4


I got the whole number side...but find it hard to get the fractional parts...after converting it to base 10, can't convert d fractional parts to the bases above..

Thank you very much

I'll answer this with a long illustration
E.g convert 7562.45 which is a decimal number to a number in base 8(octal)

To convert integer decimals to octal, divide the original number by the largest possible power of 8 and successively divide the remainders by successively smaller powers of 8 until the power is 1.

Thus we get the following with the Integer 7562

7562 / 8^4 = 1 :::: It equals 1 because 8^4 can only go into 7562 once

Therefore, 7562 - ((8^4) x 1))
=> 7562 - 4096 = 3466

Now 3466 / 8 ^ 3 = 6 :::: because 8^3 goes into 3466 six times

Therefore, 3466 - ((8^3) x 6)
=> 3466 - 3072 = 394

Now 394 / 8^2 = 6 :::: because 8^2 goes into 394 six times

Therefore, 394 / ((8^2) x 6)
=> 394 - 384 = 10

Now 10 / 8^1 = 1 :::: because 8^1 goes into 10 once only

Therefore, 10 / ((8^1) x 1)
=> 10 - 8 = 2

Now 2 / 8^0 = 2 :::: because 8^0 goes into 2 twice

Therefore, 2 / ((8^0) x 2)
=> 2-2 = 0

So if we start from the beginning:

The first attempt to divide highest power into 7562 yielded us a 1.
Next it yielded a 6
Then another 6
Then a 1
Finally a 2

Thus the octal for 7562 = 16612


=====================
Now for the Decimal Places
=====================

To convert a decimal fraction to octal, multiply by 8; the integer part of the result is the first digit of the octal fraction. Repeat the process with the fractional part of the result, until it is null or within acceptable error bounds.

0.45 x 8 = 3 + 0.6 :::: so the first yield is a 3

We take the remaining decimal now,

0.6 x 8 = 4 + 0.8 :::: so the next yeild is a 4

We take the remaining decimal now,

0.8 x 8 = 6 + 0.4 :::: so the next yield is a 6

We take the remaining decimal now,

0.4 x 8 = 3 + 0.2 :::: so next yield is a 3

OK at this point if you continue you will notice a recurring theme. If we multiple 0.2 x 8 that will leave a decimal remainder 0.6, which multiplied by 8 will leave a decimal remainder of 0.8 etcetera.

There is no point continuing so the octal of 0.45 = 0.3463

Thus the octal of 7562.45 = 16612.3463



===========================
Let's do another decimal that ends
obviously to illustrate why you stop
when its obvious you have reached
limit of the decimal octal.
===========================

0.1640625

So, as before multiple by 8 to yield a answer and continue with remaining decimals till satisfied.

Thus,

0.1640625 x 8 = 1 + 0.3125 :::: first yield = 1

0.3125 x 8 = 2 + 0.5 :::: second yield = 2

0.5 x 8 = 4 + 0.0 :::: third yield = 4

Thus 0.1640625 = 0.124 octal

If only they were all that obvious eh

OK, I hope this helps you understand how to handle the Integer and Decimals when converting to a Octal number.

Hang in there, it gets easier the more you practice.

There's another simplified way of doing it in the computer science world, I'll post that also....

Destilo:
pls kindly help me in this...@mathefaro n other maths expert in d house...

Express 110011.1001 base 2 to
1. Octal
2. Hexadecimal
3. Base 4


I got the whole number side...but find it hard to get the fractional parts...after converting it to base 10, can't convert d fractional parts to the bases above..

Thank you very much

To convert the binary number(110011.1001) to hex is easy just break it into groups of four numbers, 0011 0011 . 1001 (note that I added extra two 0s to the left to make it balanced) now with it broken into segments of four use the conversion for binary, so using the table below you get your Hex number 33.9

Hex Binary
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001
A = 1010
B = 1011
C = 1100
D = 1101
E = 1110
F = 1111

Now for octal,all you do is break it down into groups of three, your number(110011.1001) broken down would look like this 110 011 . 100 100 (note that I added extra two 0s to the right side to make it balance) then all you do is convert to octal using the table below. which you get 63.44 in octal.

Octal Binary
0 000
1 001
2 010
3 011
4 100
5 101
6 110
7 111

hope that helps.

Destilo:
1. A rough wooden block on a rough plank is drawn along the plane of the plank when inclined at 60 degree to the horizontal. Determine the coefficient of friction between two surfaces.
2. If a force of 50N moves a load of 30kg up an inclined plane. Determine the magnitude of the force if the plane is inclined at an angle of 30 degree.
Thank you

Ok..4 number 1...Frictional force(F)=coefficient f reaction(U) X normal Reaction(R)...so making coefficient of friction d subject of formulae it becomes U=F/R.... Recall F= sin@ whi

Destilo:
1. A rough wooden block on a rough plank is drawn along the plane of the plank when inclined at 60 degree to the horizontal. Determine the coefficient of friction between two surfaces.
2. If a force of 50N moves a load of 30kg up an inclined plane. Determine the magnitude of the force if the plane is inclined at an angle of 30 degree.
Thank you

Ok..4 number 1...Frictional force(F)=coefficient f reaction(U) X normal Reaction(R)...so making coefficient of friction d subject of formulae it becomes U=F/R.... Recall F= sin@ while R=cos@...so U=sin@/cos@= tan@..so d coefficient of friction is tan@=tan60= √3 so no 1 ansa is √3

chimhigher:
Ok..4 number 1...Frictional force(F)=coefficient f reaction(U) X normal Reaction(R)...so making coefficient of friction d subject of formulae it becomes U=F/R.... Recall F= sin@ while R=cos@...so U=sin@/cos@= tan@..so d coefficient of friction is tan@=tan60= √3 so no 1 ansa is √3

No 2...d magnitude wuld b Mgsin@ = 30*10*sin30=300sin30= 150N

mathefaro:
To convert the binary number(110011.1001) to hex is easy just break it into groups of four numbers, 0011 0011 . 1001 (note that I added extra two 0s to the left to make it balanced) now with it broken into segments of four use the conversion for binary, so using the table below you get your Hex number 33.9

Hex Binary
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001
A = 1010
B = 1011
C = 1100
D = 1101
E = 1110
F = 1111

Now for octal,all you do is break it down into groups of three, your number(110011.1001) broken down would look like this 110 011 . 100 100 (note that I added extra two 0s to the right side to make it balance) then all you do is convert to octal using the table below. which you get 63.44 in octal.

Octal Binary
0 000
1 001
2 010
3 011
4 100
5 101
6 110
7 111

hope that helps.

thank u very much

chimhigher:
Ok..4 number 1...Frictional force(F)=coefficient f reaction(U) X normal Reaction(R)...so making coefficient of friction d subject of formulae it becomes U=F/R.... Recall F= sin@ while R=cos@...so U=sin@/cos@= tan@..so d coefficient of friction is tan@=tan60= √3 so no 1 ansa is √3

Thnk u

chimhigher:

No 2...d magnitude wuld b Mgsin@ = 30*10*sin30=300sin30= 150N

Thnk u

Destilo:


thank u very much

you're welcome

Destilo:

Thnk u

no P.bro.Anytym

@chim and other physics expert in d house..
Help me in dis, I need to know if I got it right.thank u.

1. The average velocity of a car is 54km/h. what is the distance covered if the time taken is 10s?

2. A car starts from rest and travels at uniform acceleration until it attains a velocity of 30m/s after 6s. The distance covered is?

3. A man increased his speed by 5m/s after 30s, if his initial velocity was 20m/s. The average distance covered if his speed was at uniform acceleration would be?

4. A ball is thrown vertically upwards from the ground with an initial velocity of 50m/s. What is the total time spent by the ball in the air (g=100m/s.sq)

thank you

Destilo:
@chim and other physics expert in d house..
Help me in dis, I need to know if I got it right.thank u.

1. The average velocity of a car is 54km/h. what is the distance covered if the time taken is 10s?

First and foremost, convert the average velocity to m/s by multiplying by 5/18
==> v = 54 * 5/18 = 15m/s;
Then recall that distance = average velocity * time taken i.e. S = vt
S = ?; v = 15m/s; t = 10s
Therefore, S = 15 * 10 = 150m

2. A car starts from rest and travels at uniform acceleration until it attains a velocity of 30m/s after 6s. The distance covered is?

Since the car started from rest, initial velocity, u = 0, v = 30m/s; t = 6a; a = v/t = 30/6 = 5m/s²; S = ?
So with the available parameters, S, u, a and t, the appropriate equation to use is S = ut + (1/2)*at²;
Therefore, S = 0(6) + 0.5*5*36
S = 0+ 90
S = 90m

3. A man increased his speed by 5m/s after 30s, if his initial velocity was 20m/s. The average distance covered if his speed was at uniform acceleration would be?

V - U = 5m/s; t = 30s; U = 20m/s;
a = (V - U)/t = 5/30 = 1/6m/s²
S =ut + ½at²
S = 20(30) + ½*1/6*30*30
==> S =600 + 75
==> S = 675m

4. A ball is thrown vertically upwards from the ground with an initial velocity of 50m/s. What is the total time spent by the ball in the air (g=10m/s.sq)

thank you

u = 50m/s; v = 0; t = ?, g = 10m/s²
=> V = u - gt (minus because the body is going up against gravity)
=> 0 = 50 - 10t
=> 10t = 50
=> t = 5s

Good morning lovely people

mathefaro:
First and foremost, convert the average velocity to m/s by multiplying by 5/18
==> v = 54 * 5/18 = 15m/s;
Then recall that distance = average velocity * time taken i.e. S = vt
S = ?; v = 15m/s; t = 10s
Therefore, S = 15 * 10 = 150m


Since the car started from rest, initial velocity, u = 0, v = 30m/s; t = 6a; a = v/t = 30/6 = 5m/s²; S = ?
So with the available parameters, S, u, a and t, the appropriate equation to use is S = ut + (1/2)*at²;
Therefore, S = 0(6) + 0.5*5*36
S = 0+ 90
S = 90m


V - U = 5m/s; t = 30s; U = 20m/s;
a = (V - U)/t = 5/30 = 1/6m/s²
S =ut + ½at²
S = 20(30) + ½*1/6*30*30
==> S =600 + 75
==> S = 675m

u = 50m/s; v = 0; t = ?, g = 10m/s²
=> V = u - gt (minus because the body is going up against gravity)
=> 0 = 50 - 10t
=> 10t = 50
=> t = 5s

Good morning lovely people

Wao...u re gud
thank u

1. A car travelling at 50m/s from rest covers a distance of 10km in 40min. Calculate the acceleration.

Destilo:


Wao...u re gud
thank u

you're welcome

Destilo:



1. A car travelling at 50m/s from rest covers a distance of 10km in 40min. Calculate the acceleration.

the question is kinda wrong, I'm guessing probably with the time or with the initial velocity, because even without accelerating, if the car continue to move with speed, 50m/s, it'll exceed a distance of 10km in 40mins

Hey my people how una dey

Physics update[code][/code]

1. How much work is needed to stop a 20g bullet moving with a speed of 150m/s
2. An object of mass 10kg is placed on an inclined plane at 30degree to the horizontal. Calculate the reaction between the two surfaces (g=10m/s^2)
3. A 5kg mass on a horizontal platform accelerated at the rate of 0.1m/s^2 when a horizontal force of 10N is applied to it. Calculate the coefficient of friction between it and the platform (g=10m/s^2)

hey guys dont tink dat i fail to keep my promise by not updating after writing few pages today everytin wiped out