₦airaland Forum

overcomer90:
3. A boy looks at the image of an object in a plane mirror. He observes two images, a main bright one and the other faint. The observe image result from (A). reflection only (B). refraction only (C). diffraction and interference (D). reflection and refraction

A

Dellybanky:
A body rolls down a slope from a height of 100m. Its velocity at the foot of the slope is 20m/s. What is the percentage of its potential energy when converted into kinetic energy?

height of slope h=100m, velocity of the body v=20m/s, g=10m/s^2...P.E=mgh >>> P.E=m*10*100 >>> P.E=1000m J...K.E=1/2mv^2 >>> K.E=1/2*m*20^2 >>> K.E=1/2*m*400 >>> K.E=200m J....Energy dissipitated when the potential energy is converted to kinetic energy=1000m J - 200m J=800m J....% of P.E when converted to K.E=800m/1000m*100 >>> % of P.E converted=800/1000*100 ( m will cancel out) >>> % P.E=0.8*100=80%...therefore % of P.E when converted to K.E=80%.

@DrBESTJC ANSWERS TO THE BIOLOGY QUESTIONS (1) A (2) C (3) D (4) C (5) A (6) A (7) D (8 ) A (9) D (10) C (11) A

Sofia01:
v=0 at maximum height
v=u- at
0=u -10*3
0=u-30
u=30m/s
: to get the maximum height
v'2=u'2 -2aS
(0*0)=(30*30)-2*10*S
0=900-20S
-900=-20S
divide by -20
S=45m
: distance above window =45-5
=40m

This was what I also got as the answer but it is wrong. Thanks for even attempting the question. TeflonBlixx get the answer.

TeflonBlixx:
Maximum height = 1/2 x g x t^2
= 1/2 x 10 x 1.5 x 1.5
= 11.25
Maximum height - Height of the window
11.25 - 5
= 6.25m above the window

Thanks bro. I really appreciate.

A boy observes a cricket ball move up and down past a window 5m high. If the total time of the ball in sight is 3s. Find the height above the window that the ball rises. PLS SHOW YOUR WORKINGS!!

freshsteph:
if it's correct, kindly send the question again.....

Your answers are wrong.

Eben331:
A rope 5m long is fastened to two hooks 4m apart on a horizontal ceiling. To the rope is attached a 10kg mass so that the segments of the rope are 3m and 2m. Compute the tension in each segment. PLS SHOW YOUR WORKINGS!!

freshsteph, this is the question.

A rope 5m long is fastened to two hooks 4m apart on a horizontal ceiling. To the rope is attached a 10kg mass so that the segments of the rope are 3m and 2m. Compute the tension in each segment. PLS SHOW YOUR WORKINGS!!

freshsteph:
yah i'm back.....


eben repost your question......

Thanks bro...it has been solved already.

freshsteph:
62.5N and 37.5 N

Can you pls show ur workings??

FrancisSpecter:
September 14th/15th,like you said, just an idea.

How sure are you??

DrBESTJC:
No be small confusion be this.....where are the grammarians here ?.....Abeg explain your answer ooo...

ENGLISH

After two years of courtship,Jide is finally .......... Ngozi next week.
A) being married to
B) marrying
C) married to
D) being married by

B - Marrying. When primary auxiliaries are used to indicate future action (next week as in this case), a '-ing' verb must follow such primary auxiliary ('is' as in this case).

Pls, what is the website of eksu admission portal? Is it possible for someone to make changes after he/she has already registered? Thanks in anticipation.

Gracy3535:
dis is d method i use
Uv=Vv
Uh=Vh
10=30
10=Vh
Vh=30*10/10
Vh=30m/s
dats d ansa
cc eben331

Thanks ma. I had forgotten dat dis question had already been solved by oga samuelizz on page 60.

Damayogenius18:
..This UI practice question nah and in d option m/s was there as d unit....I don't think u need anyone to tell u DAT it is the magnitude of d final velocity they were talking abt...U should also know DAT d body will av both horizontal component and vertical component of its final velocity has it hit the ground in the second motion... V =√ Vx² + Vy²....for the first case.... H = (V² - U²)/2g..
H = (30² -10²)/20 ...H= 40m
For second motion... D height is d same....ur Vx= 10ms-¹ DAT does not change throughout d motion....u can calculate ur Vy since u know your height......Vy= 0² + 2gh.no initial vertical velocity..Vy²=2gh...Vy²= 2× 10× 40 = 800m/s
V=√ Vy² + Vx²...Vy² =800
V = √ 10² + 800
V = √100 + 800
V = √ 900
V= 30m/s....

Thanks boss. I have really forgotten dat dis question had been solved by oga samuelizz on page 60.

@DrBESTJC, ok, no problem...stop whinning me, who be boss?? I think you are right, the question is supposed to be 'what is the magnitude of the final velocity of the stone just before it hits the ground?' But that is how I dey see the question.

afoobabs:
hi everyone.
its been a long time in here

Boss, welcome back! We really need you here for English.

esivue007:
i have man,d reply ws dat "we are working on it".

You just have to wait then.

DrBESTJC, u dnt answer questions nowadays, why?? You just dey view without commenting. Help me with my question.

esivue007:
just tried it now,switched off.

Have you tried sending a message to d email??

esivue007:
ok then if u insist its 30m/s im gonna edit it(my calculations)

Bro, how far abt the man??

Dikachi483:
The answer is 30m/s.
I don't think you need any calculation for this.
The height of the cliff isn't given, and assuming it is the same stone and physical factors when he throws the stone vertically and horizontally, thus since the height, mass of stone and the initial speed is constant, then the velocity on reaching the ground will also be constant. That is 30m/s

Thanks bro...can you pls explain more??

esivue007:
who got 3om/s?

I mean 30m/s is the correct ans bt i dnt knw how they come abt it.

Eben331:
A Physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10m/s. The instant before the stone hits the ground below, it is travelling at a speed of 30m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10m/s, what is the magnitude of the stone just before it hits the ground? (a) 10m/s (b) 20m/s (c) 30m/s (d) 40m/s. PLS SHOW UR WORKINGS!!

Pls guys I need solution to this question. Where are you guys?? Cc: DrBESTJC, TeflonBlixx, Damayogenius18, hardebaryor, BiafranDel, freshsteph, Aceed, Gracy3535, Orezy5 and others. You guys should help now. I dey beg ni oo

esivue007:
40,the no was switched off.

The ans they got was 30m/s...why don't you call the chairman's phone no then?? You should also send message to the email.

esivue007:
not really cuz of dz z a symbian phone

ok, what did u get as ur ans?? Hope the man's phone no is going thru?

BiafranDel:
nah. thats not correct. The answer is c.

I think Gracy3535 and Indomixx are right...but if they are not, then show the correction to the question for us to learn.

esivue007:
im dedicatin this to eben cuz of his sympathy towards me. U=1 0 v2=30. Frm 3rd equation of motion under g =v2=u2+2gs=s=v2-u2/2a =30*30-10*10/2*10=900-100/20=800/20=40m tada!!! I reli wish i av reg

Can you pls make the workings more explicit?

esivue007:
guys, i wish to contribute my own quota too by solvin some posted question frm ya all,bt i havent been able 2 reg up till today,av did coc d day ui reg started,right now im depress a little,anybody with useful info?

Bro, don't be depressed, everything is gonna be fine. This is the helpdesk contact-08147031889 (Mr Kolade Niyi), email -ugadmission@ui.edu.ng I will suggest that you should call the phone no instead of sending message to the email because it may take a long time before ur message is being attended to.

BiafranDel:
A galvanometer has a resistance
of 5Ω. By using a shunt wire of
resistance 0.05Ω, the
galvanometer could be converted
to an ammeter capable of reading
2Amp. What is the current through the galvanometer? A. 2mA B. 10mA C. 20mA D. 25mA

Resistance across shunt = 0.05 ohms, resistance of galvanometer = 5 ohms...let current of galvanometer = IA...current flowing through the shunt = (2-I)A. >>> p.d of galvanometer = p.d across shunt >>> I*5 = 0.05(2-I) >>> 5I = 0.1-0.05I >>> 5I+0.05I = 0.1 >>> 5.01*I = 0.1 >>> I = 0.1/5.01 >>> I = 0.0199A >>> I = 19.9mA >>> I = 20mA (approx)...I don't knw whether the ans is correct or not.

Eben331:
A Physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10m/s. The instant before the stone hits the ground below, it is travelling at a speed of 30m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10m/s, what is the magnitude of the stone just before it hits the ground? (a) 10m/s (b) 20m/s (c) 30m/s (d) 40m/s. PLS SHOW UR WORKINGS!!

Someone should pls help me solve dis question.