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you are Wrong here! I presumeView attachment 38071
i dunno i just typed the answer
View attachment 38071
i) just find the area of this shaded cylindrical region
ii) 1 cubic cm = 1 ml
and 25000 l = 25000000 ml
so time taken for 1 ml (cubic cm) to come out = 1 sec
thus time taken for 25000000 ml = 25000000/(Volume u found in 'i')
Wait!!!ok il see
Bi is still nt clear by d way
ok il see
Bi is still nt clear by d way
No, You are right in Seconds, just convert to minsyou are Wrong here! I presume
I said Pi * Speed * Raduis^2ok il see
Bi is still nt clear by d way
Paen kera software haiView attachment 38071
i dunno i just typed the answer
View attachment 38071
i) just find the area of this shaded cylindrical region
ii) 1 cubic cm = 1 ml
and 25000 l = 25000000 ml
so time taken for 1 ml (cubic cm) to come out = 1 sec
thus time taken for 25000000 ml = 25000000/(Volume u found in 'i')
software?Paen kera software hai
I never understand, what is so confusing in this? ZohaKhan
u really got that?got u Sorry sorry
Oh! Finally, Useless me came Usefullgot u Sorry sorry
as we know thatCan someone tell me how to do Q25
(a)the area under the speed-time graph is the distance traveled by the object.
for any pair of triangles, u should take:For two similar triangles like these how do I determine the unknown sides? Tha angle at B is a right angle and so is ADB a right angle
AD= q cm
AB=p cm
AL= q+9 cm
BL=15 cm
BD=12 cm
DL=9 cm
please reply fast
View attachment 38213
lolX! Man I got it atleastas we know that
(a)
View attachment 38210
so in first 20 seconds the car travels D meters. that means, 0.5 * u * 20 = D (by applying the formula for area of a triangle.
10u = D
for the time after '20 seconds' we need to find when does the car travels another D meters.
so t * u = D
t * u = 10u
so we divide both sides of equation by u and we get, t = 10
thus 2D meters are traveled in 20 + 10 = 30 seconds
(b)
acceleration from 0 to 20 seconds = (u-0)/(20-0) = u/20
thus if we change acceleration to deceleration (negative acceleration: in which the speed slows down) we just change its sign i.e.
(-u/20) * 0.5 (coz Q says deceleration was half of it) = -u/40
so according to the formula acc = v/t
we get t = v/acc by dividing both sides by acc and then multiplying them with t.
so t = 60 + [(u/4) - u] / -u/40
where [u/4] - u = velocity change from 60 seconds to t seconds
hope you got it
as we know that
(a)
View attachment 38210
so in first 20 seconds the car travels D meters. that means, 0.5 * u * 20 = D (by applying the formula for area of a triangle.
10u = D
for the time after '20 seconds' we need to find when does the car travels another D meters.
so t * u = D
t * u = 10u
so we divide both sides of equation by u and we get, t = 10
thus 2D meters are traveled in 20 + 10 = 30 seconds
(b)
acceleration from 0 to 20 seconds = (u-0)/(20-0) = u/20
thus if we change acceleration to deceleration (negative acceleration: in which the speed slows down) we just change its sign i.e.
(-u/20) * 0.5 (coz Q says deceleration was half of it) = -u/40
so according to the formula acc = v/t
we get t = v/acc by dividing both sides by acc and then multiplying them with t.
so t = 60 + [(u/4) - u] / -u/40
where [u/4] - u = velocity change from 60 seconds to t seconds
hope you got it
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