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1) A car moving with speed pf 50kmph,can be stopped by applying brakes after travelling at least distance of 16m.If the same car is moving at a speed of 100kmph,the minimum distance covered by the car before stopping is
a)26m
b)24m
c)20m
d)27m
2)A stone dropped from a top of a tower is found to travel 5/9 of the height of the tower during the last second of its fall.The time fall is
a)3s
b)2s
c)4s
d)3.5s
3)A body falls from 8s from the rest if the acceleration due to gravity of earth ceases to act,the distance travelled by it during last second of it's fall is,(g=m/s sq.)
a)200m
b)160m
c)180m
d)120m
4)A stone freely falls from rest and the distance covered by it in the last second of its motion is equal to
a)4s
b)5s
c)3s
d)6s
5)a body falls freely from a height for 5 sec.Its average velocity is(g=9.8m/s sq.)
a)20.5m/s
b)24.5m/s
c)30m/s
d)49m/s
6) A stone is dropped from the 16th storey of the multistoried building and it reaches the ground in 4s. the number of stories covered by the stone in 2nd second is
a)2
b)4
c)5
d)3
I need all these answers with full explanation.Thank You!
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Answers
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There seem to be some errors in the given questions like data missing or options are wrong.
Q1. s = (v² - u²) / 2 a.
v = 0 and a = deceleration is same in both cases.
So s = - (1/2a) u²
So s2/s1 = u2² /u1²
s2 = 16 m * (100/50)² = 64 m
If s1 = 6 meters instead of 16m, then answer will be 24 meters.
Q2. h = height of tower = u t + 1/2 g t² = 0 + 1/2 g n²
Sn = distance traveled in nth sec. = u + (n-1/2) g = 0 + (n-1/2) g
Given Sn = 5/9 * h
n g - g/2 = 5 g n² /18
5 n² - 18 n + 9 = 0
(5n - 3) (n - 3) = 0 => n = 3 sec. or 3/5 sec.
Q3. t = 8 sec. u = 0. g = 10 m/s²
s = u t + 1/2 g t² = 0 + 1/2 * 10 * 8² = 320 meters
Distance traveled in 8th second = Sn = u + (n-1/2) g
= 0 + (8-1/2)* 10 = 75 meters.
Q4. The question is not correct. Data is not given.
Formula Sn = (n - 1/2) g, n = total number of seconds to fall.
Q5. u = 0.
v = u + g t = 0 + 9.8 * 5 = 49 m/sec
Average velocity/speed = (u+v)/2 = 24.5
Q6. height h = u t + 1/2 g t² . u = 0.
h = 0 + 1/2 * g * 4² = 8 g meters.
h = 8 g = 16 storeys.
So each storey has a height of g/2 meters.
Distance fallen in the 2nd second is = S_2 = u + (2 -1/2) g
= 0 + 3/2 * g meters.
= 3g/2 /(g/2) = 3 storeys.
Q1. s = (v² - u²) / 2 a.
v = 0 and a = deceleration is same in both cases.
So s = - (1/2a) u²
So s2/s1 = u2² /u1²
s2 = 16 m * (100/50)² = 64 m
If s1 = 6 meters instead of 16m, then answer will be 24 meters.
Q2. h = height of tower = u t + 1/2 g t² = 0 + 1/2 g n²
Sn = distance traveled in nth sec. = u + (n-1/2) g = 0 + (n-1/2) g
Given Sn = 5/9 * h
n g - g/2 = 5 g n² /18
5 n² - 18 n + 9 = 0
(5n - 3) (n - 3) = 0 => n = 3 sec. or 3/5 sec.
Q3. t = 8 sec. u = 0. g = 10 m/s²
s = u t + 1/2 g t² = 0 + 1/2 * 10 * 8² = 320 meters
Distance traveled in 8th second = Sn = u + (n-1/2) g
= 0 + (8-1/2)* 10 = 75 meters.
Q4. The question is not correct. Data is not given.
Formula Sn = (n - 1/2) g, n = total number of seconds to fall.
Q5. u = 0.
v = u + g t = 0 + 9.8 * 5 = 49 m/sec
Average velocity/speed = (u+v)/2 = 24.5
Q6. height h = u t + 1/2 g t² . u = 0.
h = 0 + 1/2 * g * 4² = 8 g meters.
h = 8 g = 16 storeys.
So each storey has a height of g/2 meters.
Distance fallen in the 2nd second is = S_2 = u + (2 -1/2) g
= 0 + 3/2 * g meters.
= 3g/2 /(g/2) = 3 storeys.
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