VEVEDIHNO:
A body at Rest Is Given An Uniform Acceleration Of 8m/s2 For 30s,After which the Acceleration Is Reduced To 5ms/2 for The Next 20s.The Body Maintains the Speed attained for 60s after which it is brought to rest in 20s.

(3)using the graph calculate the total disatance retardation and The Average speed.

Plz Help Me Out...Av Skecthed Out D Graph..And Also calaculated d Maximum Velocity Of The Body Plz Help Me out with d Solution On D 3rd Question dey asked.

Tanks in Advance.

Total distance = 14,840m. check picture below

plz u guyz shuld help me with diz



An anti air craft shell is fired vertically upward with a muzzle velocity of 1km/s


(a)calculate the maximum height it can attain
(b)the time taken to reach this height
(c)the instantenous velocity at the end of 20s and 50s
(d)when will it height be 37.5km


tanks in advance

VEVEDIHNO:
plz u guyz shuld help me with diz



An anti air craft shell is fired vertically upward with a muzzle velocity of 1km/s


(a)calculate the maximum height it can attain
(b)the time taken to reach this height
(c)the instantenous velocity at the end of 20s and 50s
(d)when will it height be 37.5km


tanks in advance

Using the equation of motion ''V^2=U^2+2ah, at max.height V=0 U^2=1km/s(1000x1000m/s),a=-10 since its moving upwards[when moving downwards a is +],h=?
1).0^2=1000^2+2(-10)h
0=1000000-20h
20h=1000000
h=1000000/20
h=50000m
h=50000/1000
h=50km

2)v=u-at
0=1000-10t
10t=1000
t=1000/10
t=100s
.:.t=100s
3). Inst.velocity v=u+at
v=1000-10(20)=1000-200=800m/s
v=u+at
v=1000-10(50)=1000-500=500m/s
4).h=37.5x1000=37500m
s=ut+1/2gt^2
37500=1/210t^2
37500=5t^2
t^2=37500/5
t^2=7500
t=√7500
t=86.6s

VEVEDIHNO:
plz u guyz shuld help me with diz



An anti air craft shell is fired vertically upward with a muzzle velocity of 1km/s


(a)calculate the maximum height it can attain
(b)the time taken to reach this height
(c)the instantenous velocity at the end of 20s and 50s
(d)when will it height be 37.5km


tanks in advance

(a) From v2=u2+2gh
U, initial velocity=1km/s=1000m/s
V=0m/s, g= -10m/s2( upward vertical motion), h= ?
Therefore v2-u2=-2gh
Substituting all values, we have
0-(1000)^2= -2(10)h
-(1000)^2= -20h
H= - (1000000)/ -(20) = 50000m
Maximum height=50000m
(b) from V=U+gt
V-U=gt. Where g still remains -10m/s2 and U=1000m/s
0-1000= -10t
T= -(1000)/ -(10). T=100secs
time taken to attain maximum height=100s
(c) Instantaneous velocity, V=u+gt
i. When t=20s,
V=1000+(-10)x 20 = 1000-200= 800m/s
ii. When t=50s
V=1000+(-10)x 50= 1000-500= 500m/s
(d) from V2=U2+2gh
when H,height= 37.5km or 37500m
H= u2/2g
37500= u2/2(10),
U2=37500x20=750000
U= root750000= 866.03m/s
Time= u/g
Time=866.03/ 10= 86.6secs
Therefore when the height is 37500m, the velocity would be 866m(approximately) and time would be 86.6secs.



To save yourself of all the torture of deriving those formulas, use the following formulas
Maximum height, Hmax= U2/2g
Time taken, t= U/g
Time of flight= 2u/g

gameboi:

(a) From v2=u2+2gh
U, initial velocity=1km/s=1000m/s
V=0m/s, g= -10m/s2( upward vertical motion), h= ?
Therefore v2-u2=-2gh
Substituting all values, we have
0-(1000)^2= -2(10)h
-(1000)^2= -20h
H= - (1000000)/ -(20) = 50000m
Maximum height=50000m
(b) from V=U+gt
V-U=gt. Where g still remains -10m/s2 and U=1000m/s
0-1000= -10t
T= -(1000)/ -(10). T=100secs
time taken to attain maximum height=100s
(c) Instantaneous velocity, V=u+gt
i. When t=20s,
V=1000+(-10)x 20 = 1000-200= 800m/s
ii. When t=50s
V=1000+(-10)x 50= 1000-500= 500m/s
(d) from V2=U2+2gh
when H,height= 37.5km or 37500m
H= u2/2g
37500= u2/2(10),
U2=37500x20=750000
U= root750000= 866.03m/s
Time= u/g
Time=866.03/ 10= 86.6secs
Therefore when the height is 37500m, the velocity would be 866m(approximately) and time would be 86.6secs.



To save yourself of all the torture of deriving those formulas, use the following formulas
Maximum height, Hmax= U2/2g
Time taken, t= U/g
Time of flight= 2u/g

modify ur post. Maximum Height is =U^2/2g not U2

shakrullah:

modify ur post. Maximum Height is =U^2/2g not U2

Boss, Seriously ?
Apart from that, were the solvings correct?

gameboi:
Boss, Seriously ?
Apart from that, were the solvings correct?

100% correct

Sum1 shuld assist on this question

A alloy of mass 588g and a volume 100cm^3 is made of density 8.0g/cm^3 and alluminium of density 2.7g/cm^3.
Calculate the proportion;

(i)By Volume
(ii)By Mass of the constituents of the alloy.

Fankz in advance.

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VEVEDIHNO:
Sum1 shuld assist on this question

A alloy of mass 588g and a volume 100cm^3 is made of density 8.0g/cm^3 and alluminium of density 2.7g/cm^3.
Calculate the proportion;

(i)By Volume
(ii)By Mass of the constituents of the alloy.

Fankz in advance.

let mass of Aluminum=X
p=m/v
v=m/p
100=588-X/8 + X/2.7
100x8x2.7=8X+2.7(588-X)
2160=8X+1587.6-2.7X
2160-1587.6=8X-2.7X
572.4=5.3X
X=572.4/5.3
X=108g
mass of Al=108g


but p=m/v
v=108/2.7
v=40cm^3

Kunlexic:
here is d question

A density bottle holds 250g of liqid at 30•C and only 248.5g at 60•C. Find
a.the apparent and
b.the real cubic expansivity of the liquid,if the linear expansivity of the material of the bottle is @g, let @ be alfa

Soln
¥a is the apparent cubic expansivity and the formula for ¥a=volume of liquid lost or expelled/volume of liquid remaining x temperature rise
And mass is proportional to volume.
¥a,apparent cubic expansivity=mass lost or expelled/remaining mass x rise in temperature (¤)

The temperature rises from 30•C to 60•C and the mass changes from 250g to 248.5g dat is 1.5g of liquid is lost
And change in ¤=60-30=30
¥a=1.5/248.5 x 30=0.0002012K^-1

B. The real expansivity is ¥r,cubic expansivity is ¥ and apparent expansity is ¥a,so ¥=¥r-¥a, and ¥ is 3 times the linear exapnsion and u are given the linear expansion to be 0.000006K^-1
¥=3@=3 x 0.000006=0.000018
From the formula,¥=¥r-¥a
0.000018=¥r-0.0002012
¥r=0.0002192K^-1

Hello genius..Please can you help with this? Im having difficulty solving it especially 2.

A steel is 8m long and 4mm in diameter at 0°c, calculate

1. The length of the wire at 60°c

2 the cross sectional area at 100°c

3 the increase in volume of the wire at 70°c

Given that the linear expansivity of the material of the wire is 1.2* 10^–5 K^–1

Thanks.

pls guyz help me out on dis question...

A glass bottle full of mercury has mass 5OOg. On heating through 35c of mercury are expelled. Calculate d mass of mercury remaining in d bottle. Cubic expansivity of mercury is 1.8x1O raise to power 4. Linear expansivity of glass is 8.Ox1O raise to power 6

my first assignment here is are;
( A stone is trow veritically upwaed from the ground with a velocity of 40m/s calculate)
a. the maximum high reached.
b. the time taken to attain the maximum height.
c. the time to reach the ground.
d. the velocity reached half way to the maximum height.

the second question; an inclined plane 6meter has one end raise by two meter if the efficiency of this machine is 72%
a.what is the minimum effort require to raise the load of 100N of the plane.
b. what will be the effort if the plane is perfectly smooth.

please helps my nairalander gurus

CustomII:
my first assignment here is are;
( A stone is trow veritically upwaed from the ground with a velocity of 40m/s calculate)
a. the maximum high reached.
b. the time taken to attain the maximum height.
c. the time to reach the ground.
d. the velocity reached half way to the maximum height.

the second question; an inclined plane 6meter has one end raise by two meter if the efficiency of this machine is 72%
a.what is the minimum effort require to raise the load of 100N of the plane.
b. what will be the effort if the plane is perfectly smooth.