A motorist driving at 25 meters/second decelerates to

Physics

2 answers:

prisoha [69]3 years ago

6 0

Answer:

Kinetic energy = (1/2) (mass) (speed)²

Before slowing down, the car's speed is 25 m/s,

and its kinetic energy is ...

(1/2) (1,500 kg) (25 m/s)²

= (1/2) (1,500 kg) (625 m²/s²)

= 468,750 joules .

After slowing down, the car's speed is 15 m/s,

(1/2) (1,500 kg) (15 m/s)²

= (1/2) (1,500 kg) (225 m²/s²)

= 168,750 joules.

The car lost (468,750 - 168,750) = 300,000 joules

and you heard it from the KING

timama [110]3 years ago

5 0

Kinetic energy = (1/2) (mass) (speed)²

Before slowing down, the car's speed is 25 m/s,
and its kinetic energy is ...

   (1/2) (1,500 kg) (25 m/s)²

      = (1/2) (1,500 kg) (625 m²/s²)

      =    468,750 joules .

After slowing down, the car's speed is 15 m/s,
and its kinetic energy is ...

   (1/2) (1,500 kg) (15 m/s)²

      = (1/2) (1,500 kg) (225 m²/s²)

      = 168,750 joules.

The car lost (468,750 - 168,750) = 300,000 joules of K.E.

The law of Conservation of Energy says:

That 300,000 joules had to go somewhere.

If it's a standard, gas-powered car, then the kinetic energy got
put into the brakes. The energy turned into heat, and the heat
was carried off in the air.

If it's a more modern electric or hybrid car, then the kinetic energy
spun the wheel motors, turning them temporarily into electrical
generators. The generators converted the kinetic energy into
electrical energy, which got put back into the car's batteries, and
could be used again. That's why electric cars use less gas.

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<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>

$\color{hotpink} \bold{9 \: J} \\$

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

$\underline{ \boxed{ \tt KE = \frac{1}{2} m{v}^{2} } }$

where,

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KE is the kinetic energy in J (joules)
m is the mass in kg

— — —

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KE (Kinetic energy) = ? (unknown)
m (mass) = 2 kg
v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

$\: \: \tt KE = \frac{1}{2} m{v}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: KE = \frac{1}{2} \times 2 \times {(3)}^{2} \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: KE = \frac{1}{2} \times(2\times 9) \\ \tt KE = \underline{ \boxed{ \blue{ \tt9 \: J}}}$