hmmm...try again

horpeyemmi66:

Since PE=KE, then
KE=1/2mv^2=PE=mgh
KE= 65 x10 x 50
KE= 32,500J

you didn't consider the fact that the ball has only traveled for 5s, the height Won't be 50m. Anyway, second half I'd about to begin. See you guys later

K.E=1/2mv^2
K.E= 1/2m(gt)^2
K.E=1/2 x 65 x (10 x 5)^2
K.E= 81,250J

yea,i nid more of ur physics questn@ope

Solve the last unsolved question I posted then we can continue from there...

got 69.67 bt nt sure

A BODY SLIDES DOWN AN INCLINED PLANE WHICH MAKES ANGLE OF 30 WITH THE HORIZONTAL,NEGLECTING FRICTION.CALCULATE:
a.the velocity of the body after sliding 10m from rest
b.the time taken to slide the 10m distance


MORNING TO Y'ALL AND HAPPY SUNDAY

OGOJNR:
A BODY SLIDES DOWN AN INCLINED PLANE WHICH MAKES ANGLE OF 30 WITH THE HORIZONTAL,NEGLECTING FRICTION.CALCULATE:
a.the velocity of the body after sliding 10m from rest
b.the time taken to slide the 10m distance


MORNING TO Y'ALL AND HAPPY SUNDAY

Good morning to you too...

Ans to your questions
F - mgsinΘ - Fr = 0
F - mgsinΘ - 0 = 0
F - mgsinΘ = 0
F = mgsinΘ
ma = mgsinΘ
a = gsinΘ
a = 10 x sin 30°
a = 5ms^-2

Then Velocity can be found thus,
V^2 = U^2 + 2as
V^2 = 0 + 2 x 5 x 10
V^2 = 100
V = 10m/s

Consequently,
S = Ut + 1/2at^2
S = 0 + 1/2at^2
10 = 1/2 x 5 x t^2
20/5 = t^2
t^2 = 4
t = 2s
Its an Immaculate Sunday....

Proudly C.A.C cum Baptist.

I see you @Oluwatobiloba...
--------------------------------------------------------
*PHYSICS | EXPANSIVITY

1.A triangular plate of height 6cm and base length of 12cm is made of metal of linear expansivity 3 x 10^-5 K^-1 as the plate is heated from 15°C to 70°C, the area of one side of the plate will increase to?

2. A steel plug has a diameter of 5cm at 30°C. At what temperature will it fit exactly into a hole of constant diameter 4.997cm?[Co-efficient of linear expansion of steel is 11 x 10^-6 c^-1]

Cc DrHost, OGOJNR, POlymath, Bhuuumhite, mathefaro, medwhite, funkysilver, Pr0ton, Augster, Horlarmancy, becborn, Luukasz,Francistony, AleXis0r, thaotech, atakitisam, UMartins1, darrytoz et al

horpeyemmi66:

*PHYSICS | EXPANSIVITY

1.A triangular plate of height 6cm and base length of 12cm is made of metal of linear expansivity 3 x 10^-5 K^-1 as the plate is heated from 15°C to 70°C, the area of one side of the plate will increase to?

A1 = 0.5*12*6 = 36cm^2
Superficial expansivity, B = 2æ
B = 2*3*10^-5
= 6 *10^-5K^-1
∆@= 70 -15 = 55°C
A2=A1(B∆@ + 1)
A2 = 36(6*10^-5*55 + 1)
A2 = 36(0.0033 + 1)
A2 = 36(1.0033)
A2 ~= 36.12cm^2 to 2d.p

mathefaro:

A1 = 0.5*12*6 = 36cm^2
Superficial expansivity, B = 2æ
B = 2*3*10^-5
= 6 *10^-5K^-1
∆@= 70 -15 = 55°C
A2=A1(B∆@ + 1)
A2 = 36(6*10^-5*55 + 1)
A2 = 36(0.0033 + 1)
A2 = 36(1.0033)
A2 ~= 36.12cm^2 to 2d.p

Very correct...

Am I correct @OGOJNR?

b=2a 0.00006=A2-A1/A1¤ AREA=1/2*12*6 =36CM CUBE ¤=70-15 =55CM 0.00006*36*55=A2-36 A2=36+0.1188 a2=36.1188cm sq

a=l2-l1/l1*¤ 0.000011=5-4.997/5¤ ¤=0.003/0.000055 ¤=54 bt ¤2-30=54 ¤2=54+30 =84'

yea bt can u smash dis?
A body of mass 50kg moving with a uniform speed of 5m/s comes to rest in 5 sec.calculate its acceleration before coming to rest.

OGOJNR:
yea bt can u smash dis?
A body of mass 50kg moving with a uniform speed of 5m/s comes to rest in 5 sec.calculate its acceleration before coming to rest.

You wanna play a fast one on me, but I know very well that "time is independent of mass", I guess that 50kg is just a figure-head.

1st equation of motion
V= U + at
5= 0 + at
5= 0 + 5a
a=1ms^-2

finally,i got u...that questn dnt nid any calculation since u're lukin for d acceleration before cumin to rest...wen a body undergoes uniform velocity or constant speed the acceleration is always zero...the ans is 0

hw did i do in dat expansivity questn?

OGOJNR:
yea bt can u smash dis?
A body of mass 50kg moving with a uniform speed of 5m/s comes to rest in 5 sec.calculate its acceleration before coming to rest.

You wanna play a fast one on me, but I know very well that "time is independent of mass", I guess that 50kg is just a figure-head.

1st equation of motion
V= U + at
5= 0 + at
5= 0 + 5a
a=1ms^-2

This method, I don't totally like, as the body has to travel through a distance S before coming to a halt, which is Illustrated in the second analysis

S= Ut + 1/2at^2
S= 0 + 1/2 x v/t x t^2
S= 1/2 x 5 x 5
S= 12.5m

Then,
V^2 = U^2 + 2as
5^2 = 0 + 2 x a x 12.5
25 = 25a
a= 1ms^-2...

hw did i do in dat expansivity questn

OGOJNR:
a=l2-l1/l1*¤
0.000011=5-4.997/5¤
¤=0.003/0.000055
¤=54
bt ¤2-30=54
¤2=54+30
=84'

For what question?

you are not serious o, finally I got a question of yours wrong...sure you did got me....aaaRRRGGhhhh!

You got that with 36.12 or there about as the answer very correctly.

THE 2ND ONE

OGOJNR:
THE 2ND ONE

nope, try again...use this
β= ΔA/ [A1(Θ2-Θ1)]

β= A2- A1/[A1(Θ2-Θ1)]

Cross multiplying, you have
A2 - A1 = β [ A1 ( Θ2 - Θ1 ) ]
A2 - A1 = β A1 ( Θ2 - Θ1 )
A2 = A1 + β A1 ( Θ2 - Θ1 )

After factorizing A1 out,

A2 = A1 [ 1 + β ( Θ2 - Θ1 ) ]

Recall that β = 2α,

A2= A1 [ 1 + 2α ( Θ2 - Θ 1) ]

(πd^2)/4 = (πd^2)/4 [ 1 + 2α ( Θ2 - Θ1) ]

π can cancel out up there...

Just insert your parameters, be careful though.

itz linear expansivity,hw did it become superficial,sir..i dnt undastand

Hello every 1 av bin luking at this page as a guest and I lyk the spirit here, I mean serious minds, its gooood. Pray we get admitted. Let's keep the fire burning

OGOJNR:
itz linear expansivity,hw did it become superficial,sir..i dnt undastand

The relationships between linear, area/superficial and volumic/cubic expansivity.

For Area/superficial expansivity, β= 2α

for Cubic/ Volumic, γ =3α

α= β/2

So γ is also = 3β/2

α= γ/3

β=2γ/3

...and thanks nessie17, we all appreciated that timely post.

i knw bt PLS solve it...stil got 84

OGOJNR:
itz linear expansivity,hw did it become superficial,sir..i dnt undastand

my brother, it's not linear expansivity. It's actually superficial because Its the cross sectional area of the the plug that needs to decrease to fit in into the hole and not the length. So if you calculate well, you'll get a temperature less than the given one in which the radius is 5cm. I hope it's clear now.
**Winks**
For me, I'm still at a birthday party(owo epo things) lol

no wonder,i sensed manipulation...i dnt see cross sectional area in d questn nw

OGOJNR:
i knw bt PLS solve it

Ok, here is the thing,
note the following well because once a mistake is made in them, it just sends all your workings crumbling like a pack of dominoes.

A2 = 5cm
A1 = 4.997cm
Θ2 = 30°C
Θ1 = ??

From here,

A2 - A1 = β A1 ( Θ2 - Θ1 )
A2 = A1 + βA1 ( Θ2 - Θ1 )
A2 = A1 [ 1 + β ( Θ2 - Θ1 ) ]
A2 = A1 [ 1 + 2α ( Θ2 - Θ1) ]

(πd^2)/4 = (πd^2)/4 [ 1 +( Θ2 - Θ1 )

π cancels out,

(5^2)/4= (4. 997)^2/4[ 1 + 2( 11 x 10^ -6) (30 - Θ1)]

25/4 = 24.97/4[1 +[22 x10^-6] (30 - Θ1)]

6.25/6.242 = 1 + (22 x 10^-6) (30 - Θ1)

1.0012 - 1= 22 x 10^-6 (30 - Θ1)

0.0012/ 0.000022 = 30 - Θ1

54.54 - 30 = -Θ1

-Θ1 = 24.54

→ Θ1 = - 24.54°C