Asked by srikar
1.A body with an initial velocity of 18km/hr accelerates uniformly at the rateof 9cm/sec2over a distance of 200m.Calculate, a)the acceleration in m/sec2 b)its final velocity in m/s
2.The distance travelled by a body moving with uniform acceleration in the nth second is given by Sn=3.8+0.4n,find the velocity of the body .
2.The distance travelled by a body moving with uniform acceleration in the nth second is given by Sn=3.8+0.4n,find the velocity of the body .
Answers
Answered by
Elena
1.
vₒ = 18 km/hr = 18000/3600 = 5 m/s.
a = 9 cm/s² = 0.09 m/s².
s =(v² - vₒ²)/2•a.
v = sqrt (vₒ² +2•a•s) = sqrt(25 +2•0.09•200) =7.81 m/s.
vₒ = 18 km/hr = 18000/3600 = 5 m/s.
a = 9 cm/s² = 0.09 m/s².
s =(v² - vₒ²)/2•a.
v = sqrt (vₒ² +2•a•s) = sqrt(25 +2•0.09•200) =7.81 m/s.
Answered by
Elena
2.
Distance traveled by a body in nth second is
Sn = u + (2n-1)•a/2.
If you know acceleration "a" you can find velocity "u"
Distance traveled by a body in nth second is
Sn = u + (2n-1)•a/2.
If you know acceleration "a" you can find velocity "u"
Answered by
Henry
Vo = 18km/h =18000m/3600s=5m/s=Initial
velocity.
1a. a =9cm/s^2 * 0.01m/cm = 0.09 m/s^2.
1b. V^2 = Vo^2 + 2ad
V^2 = 5^2 ++ 0.18*200 = 61
V = 7.81 m/s. = Final velocity.
velocity.
1a. a =9cm/s^2 * 0.01m/cm = 0.09 m/s^2.
1b. V^2 = Vo^2 + 2ad
V^2 = 5^2 ++ 0.18*200 = 61
V = 7.81 m/s. = Final velocity.
Answered by
saila
what is "sqrt" please can i know?????
Answered by
Elena
sqrt=square root=quadratic root=second root
Answered by
Max
What is the way to study physics
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