1.which of the following cannot be made by green plants? A. vitamins B. fats C. carbohydrates D. minerals E. proteins.

1. which of these metals, mg, fe, pb and cu will dissolve in dilute HCl? A. all the metals B. mg, fe and cu
C. mg, fe and pb
D.mg and fe only
2. The cracking process is very important in the petroleum industry bcs it.
A. gives purer products
B. yields more lubricants
C yields more engine fuels.
D. yields more asphalts
E. yields more candle wax

1. Granted you are my senior by one year, you cannot______around as if I were your servant.A. control B. dribble. C. order D. ask E. compel.
2. it was a blind alley, I could not walk any____ A.further
B. farthest.
C. furthest
D.farther.
3.Because the referee was partial and his opponent rough, the loser of the wrestling match received ____cheers than the winner.
A. many more
B.overwhelming
C a little more
D.several

vicstiqs95:
1. Granted you are my senior by one year, you cannot______around as if I were your servant.A. control B. dribble. C. order D. ask E. compel.
2. it was a blind alley, I could not walk any____ A.further
B. farthest.
C. furthest
D.farther.
3.Because the referee was partial and his opponent rough, the loser of the wrestling match received ____cheers than the winner.
A. many more
B.overwhelming
C a little more
D.several

lemme try ...C.D.A

vicstiqs95:
1. which of these metals, mg, fe, pb and cu will dissolve in dilute HCl? A. all the metals B. mg, fe and cu
C. mg, fe and pb
D.mg and fe only
2. The cracking process is very important in the petroleum industry bcs it.
A. gives purer products
B. yields more lubricants
C yields more engine fuels.
D. yields more asphalts
E. yields more candle wax

lemme try ...D.C

Orezy5:

5/6
Check number 3 again

Av checked sir... I still got same answer...11N...

Samsonyte:
correct!

ok bro

Aybalance:
Av checked sir... I still got same answer...11N...

Yeah..
I'm sorry, You're correct

I made a mistake while I was solving the question

Orezy5:
PHYSICS (EQUILIBRIUM OF FORCES)
1. A uniform beam of length 4m and mass 20kg is supported at both ends. A girl of mass 40kg stands on the beam at a distance of 1.5m from one of the supports. The reactions at the supports are
a. 150N, 250N
b. 250N, 350N
c. 200N, 300N
d. 300N, 300N


2. Weights of 0.2N and 0.5N are placed at the 30cm and 80cm marks respectively on a uniform metre rule. If the metre rule balances horizontally on a knife edge at the 60cm mark, the weight of the metre rule is
a. 0.4N
b. 0.1N
c. 3.3N
d. 0.7N


3. A uniform metre rod of mass 1.5kg is pivoted at one end. A weight of 7N is placed at the 50cm mark of the rod. The vertical force which should be applied at the other end to maintain the rod in equilibrium in the horizontal position is
a. 22N
b. 10.6N
c. 8.5N.
d. 11N


4. A mass of 10g Is attached to the 0cm mark of a light metre rule while a mass of 40g is attached to the other end. The centre of gravity of the system is located at the
a. 20cm mark
b. mid point of the metre rule
c. 60cm mark
d. 80cm mark


5. A force of 8.75N is applied to a mass of 3.5kg placed on a horizontal surface. If the coefficient of friction is 0.2, the acceleration of the mass is
a. 0.5m/s²
b. 1.0m/s²
c. 1.5m/s²
d. 3.3m/s²


7. A crate of mass 25kg moving with a speed of 3m/s on a rough horizontal floor is brought to rest after sliding a distance of 2.5m on the floor. The coefficient of sliding friction is
a. 0.09
b. 0.18
c. 0.36
d. 0.25

Pls I will love to see the correct workings for no.1 and 3. don't really understand it.

Aybalance:
lemme try ...D.C

I will like you to explain y the first 1 is D. think u r correct.

A normal good with close substitutes is likely to have its price elasticity of demand

A. between zero and one
B. equal to unity
C. greater than unity
D. less than unity

A rightward shift in the (PPF) may be due to

A. use of inferior inputs
B. ineffiency
C. improvemeny in production techniques and
practices
D. changes in the product mix

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Orezy5:
PHYSICS (EQUILIBRIUM OF FORCES)
1. A uniform beam of length 4m and mass 20kg is supported at both ends. A girl of mass 40kg stands on the beam at a distance of 1.5m from one of the supports. The reactions at the supports are
a. 150N, 250N
b. 250N, 350N
c. 200N, 300N
d. 300N, 300N


2. Weights of 0.2N and 0.5N are placed at the 30cm and 80cm marks respectively on a uniform metre rule. If the metre rule balances horizontally on a knife edge at the 60cm mark, the weight of the metre rule is
a. 0.4N
b. 0.1N
c. 3.3N
d. 0.7N


3. A uniform metre rod of mass 1.5kg is pivoted at one end. A weight of 7N is placed at the 50cm mark of the rod. The vertical force which should be applied at the other end to maintain the rod in equilibrium in the horizontal position is
a. 22N
b. 10.6N
c. 8.5N.
d. 11N


4. A mass of 10g Is attached to the 0cm mark of a light metre rule while a mass of 40g is attached to the other end. The centre of gravity of the system is located at the
a. 20cm mark
b. mid point of the metre rule
c. 60cm mark
d. 80cm mark


5. A force of 8.75N is applied to a mass of 3.5kg placed on a horizontal surface. If the coefficient of friction is 0.2, the acceleration of the mass is
a. 0.5m/s²
b. 1.0m/s²
c. 1.5m/s²
d. 3.3m/s²


7. A crate of mass 25kg moving with a speed of 3m/s on a rough horizontal floor is brought to rest after sliding a distance of 2.5m on the floor. The coefficient of sliding friction is
a. 0.09
b. 0.18
c. 0.36
d. 0.25

..1...B (250N,350N)...2..A ( 0.4N)...3..A (22N)...4..D (80 cm mark)..5..A..(0.5N)...7...B (0.18)....

vicstiqs95:


Pls I will love to see the correct workings for no.1 and 3. don't really understand it.

No 1...(200 * 2) + (40 * 15) =4N.........400+600=4N... 1000=4N...N= 250....the whole weight of the plank = 600N....the upward force has to balance the downward force...so..the downward force= 600N...one of the upwar force which is the reaction = 250 N...the other reaction will = 600- 250= 350N......the concept is that you av to take your pivot at one of the reaction...where you take your pivot,the force there will be ignore or considered as Zero.........m sori...can't explain beta...

vicstiqs95:


Pls I will love to see the correct workings for no.1 and 3. don't really understand it.

...No..3.....on a uniform metre rule,the total weight lies at the centre...a uniform rule is always 100 cm or 1m.....the plank weighed 1.5kg or 15N.. The centre of 100 cm plank is 50cm....another weight of 7N is place at the centre or at 50cm...total weight is now= 15N+ 7N= 22N...we are asked the upward force that will balance the downward force....since the downward force = 22N...upward force= 22N

For others ...que 2....( 0.2 * 30)+ W * 10 =0.5 *20... 6 +10W = 10.....10 - 6 =10W.....4 =10W.....W=0.4N..

ques 4...10 * x =40(100- x)..10x =4000 -40x...10x+40x =4000....50x = 4000...x = 80cm

ques 5.... net force = ma...net force = force applied - frictional force..... So,..
fa - fr= ma..but fr = umg.....fa- umg = ma....8.75 - ( 3.5 * 0.2 *10)= 3.5a...8.75- 7 = 3.5a..1.75 =3.5a...a= 1.75/3.5...a= 0.5..

Damayogenius18:
ques 5.... net force = ma...net force = force applied - frictional force..... So,..
fa - fr= ma..but fr = umg.....fa- umg = ma....8.75 - ( 3.5 * 0.2 *10)= 3.5a...8.75- 7 = 3.5a..1.75 =3.5a...a= 1.75/3.5...a= 0.5..

what textbook do u make use of

Que 7...we can use 2 formula...but lemme simplify it...since the body is coming to rest...the total net force acting on it is only the frictional force....fnet = ma..fnet= fr..so fr= ma...let's get a first...vsquare = usquare + 2as...vsquare = 0.. Since the crate came yo rest....then, usquare = 2as..a= usquare/2s..... u is 3..while s is 2.5...a= 9/ 5 which is = 1.8...fr = ma...fr = umg...so,
umg = ma....ug = a ..u= a/g..= 1.8/ 10 ...u =0.18..

dannyboii1:




what textbook do u make use of

lamlad physics and science physics

Orezy5:
BIOLOGY


(GENETICS : SEX DETERMINATION AND SEX-LINKED CHARACTERS)


1. If a baby is a female, her mother's ovum must have been fertilized by a sperm carrying the chromosome
a. X
b. XY
c. YY
d. Y


2. In which of the following crosses will all the female offspring be colour blind?
a. colour blind mother x colour blind father
b. colour blind mother x normal father
c. carrier mother x colour blind father
d. carrier mother x normal father


3. A sex linked trait can never be transmitted from
a. mother to daughter
b. father to son
c. mother to son
d. father to daughter


4. Which one of the following characteristics is not sex linked?
a. colour blindness
b. baldness
c. haemophilia
d. muscular dystrophy
e. none of the above


5. Which of the following is true of the children of a haemophilic man who marries a woman that is not haemophilic and does not carry the trait?
a. all their sons will be haemophilic
b. all their daughters will be haemophilic
c. all their daughters will be carriers
d. all their sons will be carriers


6. Women do not suffer from colour blindness because
a. the trait is sex linked
b. only men are colour blind
c. the genes are recessive and sex linked
d. the genes occur on both the X and Y chromosomes


7. The pair of colours that cannot be distinguished by a person suffering from colour blindness is
a. white and blue
b. yellow and green
c. red and blue
d. red and green
e. yellow and red


8. At what stage in the life history of a mammal is the sex of an individual set?
a. at puberty
b. at birth
c. at conception
d. during meiosis
e. at adolescence


9. Which of the following is a diploid cell?
a. spermatogonia
b. spermatozoa
c. ovum
d. spermatids
e. none of the above

Afternoon Bosses

1..A..2..A..3..B..4..D..5..C..6...C..7..D.. 8...C..9...A

Damayogenius18:
No 1...(200 * 2) + (40 * 15) =4N.........400+600=4N... 1000=4N...N= 250....the whole weight of the plank = 600N....the upward force has to balance the downward force...so..the downward force= 600N...one of the upwar force which is the reaction = 250 N...the other reaction will = 600- 250= 350N......the concept is that you av to take your pivot at one of the reaction...where you take your pivot,the force there will be ignore or considered as Zero.........m sori...can't explain beta...

thanks bro, really appreciate.

Damayogenius18:
1..A..2..A..3..B..4..D..5..C..6...C..7..D.. 8...C..9...A

muscular dystrophy is sex- linked.

vicstiqs95:


muscular dystrophy is sex- linked.

...thanks..I don't know that....

vicstiqs95:
1.which of the following cannot be made by green plants? A. vitamins B. fats C. carbohydrates D. minerals E. proteins.

..fats

Samsonyte:
A normal good with close substitutes is likely to have its price elasticity of demand

A. between zero and one
B. equal to unity
C. greater than unity
D. less than unity

A rightward shift in the (PPF) may be due to

A. use of inferior inputs
B. ineffiency
C. improvemeny in production techniques and
practices
D. changes in the product mix

C C I guess PPF or PPC

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[quote author=Damayogenius18 post=66602874]...No..3.....on a uniform metre rule,the total weight lies at the centre...a uniform rule is always 100 cm or 1m.....the plank weighed 1.5kg or 15N.. The centre of 100 cm plank is 50cm....another weight of 7N is place at the centre or at 50cm...total weight is now= 15N+ 7N= 22N...we are asked the upward force that will balance the downward force....since the downward force = 22N...upward force= 22N[/quote...I got 11N...considering the supports at both ends.the upward force to be applied at each end is 11N...

vicstiqs95:


I will like you to explain y the first 1 is D. think u r correct.

..hydrochloric acid is hcl(g) in water...metals more reactive than hydrogen like Fe,Mg,and Pb reacts with Hcl to form salt and h2 gas...The salts dissolves in the water of the acid...Pbcl2 is insoluble in water as it accumulates around the lead and stops the reaction...so only Fe and Mg dissolve in hcl