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Katen [24]
2 years ago
14

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener

gy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.
Physics
1 answer:
AVprozaik [17]2 years ago
8 0

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

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Answer:

Lilly's speed is two times John's speed.

Explanation:

m = Mass

a = Acceleration

t = Time taken

u = Initial velocity

v = Final velocity

The force they apply on each other will be equal

F=ma\\\Rightarrow a_l=\frac{F}{m_l}

F=ma\\\Rightarrow a_j=\frac{F}{2m_l}\\\Rightarrow a_j=\frac{1}{2}a_l

v=u+at\\\Rightarrow v_l=0+\frac{F}{m_l}\times t\\\Rightarrow v_l=a_lt

v=u+at\\\Rightarrow v_l=0+\frac{F}{2m_l}\times t\\\Rightarrow v_j=\frac{1}{2}a_lt\\\Rightarrow v_j=\frac{1}{2}v_l\\\Rightarrow v_l=2v_j

Hence, Lilly's speed is two times John's speed.

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3 years ago
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