On dry concrete, a car can decelerate at a rate of 7.00 m/s2 , whereas on wet concrete it can decelerate at only 5.00 m/s2 . Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.

Explanation:

a ) Let the distance required in former case be d₁ .

initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 7.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 7 x d₁

d₁ = 64.28 m

b) initial velocity u = 30 m /s , final velocity v =0 , deceleration a = 5.00 m /s²

v² = u² - 2 a s

0 = 30² - 2 x 5 x d₂

d₂ = 90  m

c)

t = .5 s

s₁  = ut - .5 at²

= 30 x .5 - .5 x 7 x .5²

= 15 - .875

= 14.125 m

t = .5 s

s₂  = ut - .5 at²

= 30 x .5 - .5 x 5 x .5²

= 15 - .625

= 14.375  m

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j