Answer:
Explanation:
Let after time t , Tina catches up David .
Distance travelled by them are equal ,
Distance travelled by Tina
s = ut + 1/2 a t²
= .5 x 2.10 t²
= 1.05 t²
Distance travelled by David
= 30 t ( because of uniform velocity )
1.05 t² = 30t
t = 28.57 s
Distance travelled by Tina
= 1/2 a t²
= .5 x 2.10 x 28.57²
= 857 m approx.
Answer:
1200 Sm^2mol^-1
Explanation:
Given data :
conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1
conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1
Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1
Resistance = 33.21 Ω
where conductivity can be expressed as = 
hence cell constant = conductivity * Resistance
= 1.0879 * 33.21 = 36.13m^-1
conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300
= 0.120 Sm^-1
<u>Determine the molar conductivity of acetic acid</u>
= ( kCH3COOH * 1000 ) / C
C = 0.1 mol dm
= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1
To determine the height of the object given the time, we simply use the given relation between height and time in the problem statement. It is given as:
h = -16t^2 + 127t
We substitute 55 seconds to t and obtain,
h = -16(55)^2 + 127(55)
h = - 41415