Question An electric heater is used to melt a block of ice of mass 1.5kg if the heater is powered by 12volts of a battery and current of 20amp flows through the coil calculate the time taken to melt the ice at 0¤C(specific latent heat of fusion of ice=336 X 10^3g/kg)

Jskelly11:
Question An electric heater is used to melt a block of ice of mass 1.5kg if the heater is powered by 12volts of a battery and current of 20amp flows through the coil calculate the time taken to melt the ice at 0¤C(specific latent heat of fusion of ice=336 X 10^3g/kg)

ivt=ml;20 X 12 X t=1.5 X 336 X 10^3
240t=504 X 10^3
t=(504 X 10^3)/240
t=2.1 X 10^3
t=2100s or 35mins

Bhuumhite:
ivt=ml;20 X 12 X t=1.5 X 336 X 10^3
240t=504 X 10^3
t=(504 X 10^3)/240
t=2.1 X 10^3
t=2100s or 35mins

Nice you can also use pt=ml

Jskelly11:
Nice you can also use pt=ml

yh,p=iv

jossayo:
During the same time interval,it is observed that a train travels the same distance as does a lorry.the two vehicles there4 have the same?

Same and time.

Another question calculate the final temperature of 250g of water at 60¤C if added to 150g of water at 5¤C specific heat capacity of water =420j/kg ₹₹₹physics is sweet₹₹₹

Jskelly11:
Another question calculate the final temperature of 250g of water at 60¤C if added to 150g of water at 5¤C specific heat capacity of water =420j/kg ₹₹₹physics is sweet₹₹₹

m1C(t1-t)=m2C(t-t2)
m1=250g=0.25kg
m2=150g=0.15kg
C=420j/kg
t1=60¤C
t2=5¤C
t=?
0.25 X 420 X (60-t)=0.15 X 420 X (t-5)
105(60-t)=63(t-5)
6300-105t=63t-315
-105t-63t=-315-6300
-168t=-6615
t=-6615/-168
t=39.38¤C

Bhuumhite:

m1C(t1-t)=m2C(t-t2)
m1=250g=0.25kg
m2=150g=0.15kg
C=420j/kg
t1=60¤C
t2=5¤C
t=?
0.25 X 420 X (60-t)=0.15 X 420 X (t-5)
105(60-t)=63(t-5)
6300-105t=63t-315
-105t-63t=-315-6300
-168t=-6615
t=-6615/-168
t=39.38¤C

good Ma are you a jambite

VEVEDIHNO:
When taking a penalty kick a footballer applies a force of 30N for a period of 0.05s.
if the mass of the ball is 0.075kg.calculate the speed with which the ball moves.

pls help me out.

The formula to use is:
v=u + at
v(final velocity) = ?
u(initial velocity) = 0m/s
t(time) = 0.05s
a(acceleration) = force/mass=30÷0.075= 400m/s^2
v=0 + 400×0.05
v=20m/s.
NOTE: the v(final velocity) is the speed requested.

VEVEDIHNO:
When taking a penalty kick a footballer applies a force of 30N for a period of 0.05s.
if the mass of the ball is 0.075kg.calculate the speed with which the ball moves.

pls help me out.

A better approace

F.T = MV - MU
Where
F = 30N
T = 0.05s
M = 0:075kg
U = 0
V = ?

30 x 0.05 = 0.075 x V - 0:075 x 0

30 x 0.05 = 0.075V - 0

V = 30 x 0.05/0:075

V = 20M/S

VEVEDIHNO:
When taking a penalty kick a footballer applies a force of 30N for a period of 0.05s.
if the mass of the ball is 0.075kg.calculate the speed with which the ball moves.

pls help me o
ut.

Impulse=momentum
Ft=mv
30 x 0.05=0.075v
V=30 x 0.05/0.075=20m/s

Jskelly11:
Question An electric heater is used to melt a block of ice of mass 1.5kg if the heater is powered by 12volts of a battery and current of 20amp flows through the coil calculate the time taken to melt the ice at 0¤C(specific latent heat of fusion of ice=336 X 10^3g/kg)

H=p x t=ivt=ml
20 x 12 x t=15 x 336 x 10^3
240t=504 x 10^3
t=504x10^3/240=2100s

Jskelly11:
Another question calculate the final temperature of 250g of water at 60¤C if added to 150g of water at 5¤C specific heat capacity of water =420j/kg ₹₹₹physics is sweet₹₹₹

Heat loss= heat gained
250 x 420(60-@)=150 x 420 (@-5)
105000(60-@)=63000(@-5)
6300000-105000@=63000@-315000
6615000/168000=168000@/168000

@=39.4°¢

Jskelly11:
good Ma are you a jambite

yes sir

mathefaro:
speed

U guyz Shuld Help Me Out Plz...

VEVEDIHNO:



U guyz Shuld Help Me Out Plz...

what help do you want from me

please help me

WHAT COURSE CAN I STUDY TO BECOME A CHEMISTRY TEACHER
Chemistry OR Education and Chemistry

And how many years course is it?

Jskelly11:
Another question calculate the final temperature of 250g of water at 60¤C if added to 150g of water at 5¤C specific heat capacity of water =420j/kg ₹₹₹physics is sweet₹₹₹

sowie sir, buh isn't specific heat capacity of water=4200j/kg

FunkySilver:

sowie sir, buh isn't specific heat capacity of water=4200j/kg

yah thanks typos tinz

plz answer this, with a well detailed explanation.
Que: An object is projected vertically upwards from a certain height, hits the ground after a few seconds. it has its maximum speed:

A. at halfway mark travelling upwards
B. at halfway mark travelling downward
C. at exactly its highest point
D. just before hitting the ground
E. just before it reaches its highest point

Preboy:
plz answer this, with a well detailed explanation.
Que: An object is projected vertically upwards from a certain height, hits the ground after a few seconds. it has its maximum speed:

A. at halfway mark travelling upwards
B. at halfway mark travelling downward
C. at exactly its highest point
D. just before hitting the ground
E. just before it reaches its highest point

C-just before hitting the ground because when a body is thrown vertically upwards, it's speed gradually decreases until it gets to the maximum height where it's speed is zero but as the body starts moving downwards, it accelerates, increasing it's speed until it gets to its maximum magnitude just before hitting the ground where the speed goes back to zero (body not moving anymore)

mathefaro:
)

Bhuumhite:

Well done for keeping here alive and solving the questions

i ave a problem wit a questn,n i dnt if it wasn't wellsoved ni. .if yu have newskool physics by anyakoha,go to page44,example 4.4,pls luk am wella,i tnk sumtin is wrong with the solutn

jossayo:
i ave a problem wit a questn,n i dnt if it wasn't wellsoved ni. .if yu have newskool physics by anyakoha,go to page44,example 4.4,pls luk am wella,i tnk sumtin is wrong with the solutn

post d qst hre instead

d question goes thus.
A density bottle holds 250g of liquid at 30 degree C and only 248.5g at 60 degree 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 Dg=0.000006 K^-1


Culled from New School Physics page 44 example 4.4

jossayo:
i ave a problem wit a questn,n i dnt if it wasn't wellsoved ni. .if yu have newskool physics by anyakoha,go to page44,example 4.4,pls luk am wella,i tnk sumtin is wrong with the solutn

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

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

dnt u tnk d temp shud be converted to kelvin,since u re gettin ur ansa in per kelvin

jossayo:
dnt u tnk d temp shud be converted to kelvin,since u re gettin ur ansa in per kelvin

no don't convert it,either per kelvin or per degree celcius

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.

(1) Calculate the maximum Velocity of the body

(2)Skecth The velocity time graph

(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.

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.

(1) Calculate the maximum Velocity of the body

(2)Skecth The velocity time graph

(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 retardation....i get -12m/s^2 and i got 110.7m/s(not sure if it's correct) for average speed..

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.

(1) Calculate the maximum Velocity of the body

(2)Skecth The velocity time graph

Tanks in Advance.

maximum Velocity v2 = 340m/s. Check picture below for details