Answer:
speed = 72.75km/hr
Explanation:
Let us first determine the Kinetic energy (KE) of the 18000kg truck
[tex]KE = \frac{1}{2} mv^2\\where:\\m = mass = 18000\\v = velocity = 21km/hr\\\therefore KE = \frac{1}{2} \times 18000 \times (21)^2\\= 9000 \times 441\\KE = 3,969,000\ Joules\\= 3,969\ KJ[/tex]
Next we will substitute this value of kinetic energy into the KE equation for the 1500kg compact car
[tex]KE = \frac{1}{2} mv^2\\3969000 = \frac{1}{2} \times 1500 \times v^2\\3969000 = 750 \times v^2\\v^2 = \frac{3969000}{750} = 5292\\v = \sqrt{5292} \\v = 72.75km/hr[/tex]
The speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h
The kinetic energy of an object is known to be the energy at work and in motion. It is usually expressed as;
[tex]\mathbf{K.E = \dfrac{1}{2}mv^2}[/tex]
To determine the speed at which the car will have the same kinetic energy as the truck, we can say that:
[tex]\mathbf{\implies \dfrac{1}{2}m_c v_c^2= \dfrac{1}{2}m_t v_t^2}[/tex]
∴
[tex]\mathbf{\implies \dfrac{1}{2}\times 1500\times v_c^2= \dfrac{1}{2}\times 18000 \times 21^2}[/tex]
[tex]\mathbf{v_c^2= \dfrac{3969000}{ 750}}[/tex]
[tex]\mathbf{v_c^2=5292}[/tex]
[tex]\mathbf{v_c=\sqrt{5292}}[/tex]
[tex]\mathbf{v_c\simeq 73 km/h}[/tex]
Therefore, we can conclude that the speed at which the 1500 kg compact car have the same kinetic energy as the truck is 73 km/h
Learn more about kinetic energy here:
https://brainly.com/question/12669551?referrer=searchResults
