contestada

The Kentucky Derby is one of the most exciting horse races run every year. In 1973, the horse Secretariat set a record that still stands. This 500-kg horse ran the 2000-m track in 120 s. What was Secretariat’s average kinetic energy? (1 kJ = 1000 J)

A. 4.2 kJ
B. 8.3 kJ
C. 69.0 kJ
D. 139.0 kJ

Respuesta :

dsdrajlin

Answer:

C. 69.0 kJ

Explanation:

Step 1: Given data

Mass of the horse (m): 500 kg

Distance traveled by the horse (d): 2000 m

Time elapsed (t): 120 s

Step 2: Calculate the speed of the horse (v)

The speed of the horse is equal to the distance traveled divided by the time elapsed.

v = d/t

v = 2000 m/120 s

v = 16.6 m/s

Step 3: Calculate the kinetic energy (K) of the horse

We will use the following expression.

K = 1/2 × m × v²

K = 1/2 × 500 kg × (16.6 m/s)²

K = 6.90 × 10⁴ J

Step 4: Convert "K" to kJ

We will use the conversion factor 1 kJ = 1000 J.

6.90 × 10⁴ J × (1 kJ/1000 J) = 69.0 kJ

The average kinetic energy is option C. 69.0 kJ

Calculation of the average kinetic energy:

But before that we have to do following calculations:

The Mass of the horse = (m)= 500 kg

The Distance traveled by the horse= (d)=  2000 m

And, Time elapsed (t) = 120 s

Now the speed of the horse (v) should be

v = d/t

= 2000 m/120 s

= 16.6 m/s

Now kinetic energy (K) of the horse should be

K = 1/2 × m × v²

K = 1/2 × 500 kg × (16.6 m/s)²

K = 6.90 × 10⁴ J

Now we have to Convert "K" to kJ

Since 1 kJ = 1000 J.

So,

= 6.90 × 10⁴ J × (1 kJ/1000 J)

= 69.0 kJ

hence, The average kinetic energy is option C. 69.0 kJ

learn more about kinetic energy here: https://brainly.com/question/13601399

Otras preguntas