a hawk is chasing a sparrow at the beginning of the chased the hawks kinetic energy is 5.1 joules at the end of the chase the hawk doubles in speed what is the best prediction for its kinetic energy at the end of the chase
20.4 j
2.55 j
5.1 j
10.2 j
10.2 j
a)20.4 j i took the test and that is the right answer. also when velocity (speed) doubles, kinetic energy quadruples.
Well, well, well. A hawk chasing a sparrow, huh? Talk about some high-speed bird drama! Now, let's get to the point. If the hawk doubles its speed during the chase, we can expect its kinetic energy to also double. So, taking into account the initial kinetic energy of 5.1 joules, the best prediction for the hawk's kinetic energy at the end of the chase would be 10.2 joules. That hawk is really going to be flying with a lot of energy!
To determine the best prediction for the hawk's kinetic energy at the end of the chase, we can use the fact that kinetic energy is proportional to the square of an object's speed.
Given that the hawk's initial kinetic energy is 5.1 joules, and it doubles its speed by the end of the chase, we can calculate the new kinetic energy as follows:
New kinetic energy = (Initial kinetic energy) * (New speed^2 / Initial speed^2)
Since the new speed is twice the initial speed, we have:
New kinetic energy = (5.1 joules) * ((2 × initial speed)^2 / (initial speed)^2)
Simplifying this equation, we get:
New kinetic energy = (5.1 joules) * (4)
Therefore, the best prediction for the hawk's kinetic energy at the end of the chase is:
20.4 joules
So, the correct answer is: 20.4 j
To predict the kinetic energy at the end of the chase for the hawk, we need to understand that kinetic energy is directly proportional to the square of the velocity.
Given that the hawk doubles in speed, we can use this information to determine the best prediction for its kinetic energy at the end of the chase.
Let's calculate the initial velocity (v1) and the final velocity (v2) based on the given information:
v1 (initial velocity) = √(2KE1/m) -- (Equation 1)
v2 (final velocity) = 2v1 -- (Equation 2)
Here, KE1 represents the initial kinetic energy of the hawk, and m represents the mass (which we assume remains constant).
From the given information, KE1 = 5.1 Joules.
Now, substituting the values in Equation 1:
v1 = √(2 * 5.1 / m) -- (Equation 3)
To predict the kinetic energy at the end of the chase, we need to find the square of v2 (final velocity) and multiply it by m (since KE2 = 0.5 * m * v2^2).
Substituting the value of v2 from Equation 2 in Equation 3:
v2 = 2 * v1
v2 = 2 * √(2 * 5.1 / m)
v2 = 2√(10.2 / m)
Now we need to square v2 and multiply it by m:
KE2 = 0.5 * m * v2^2
KE2 = 0.5 * m * (2√(10.2 / m))^2
KE2 = 0.5 * m * 4 * (10.2 / m)
KE2 = 10.2 Joules
Therefore, the best prediction for the hawk's kinetic energy at the end of the chase is 10.2 Joules.
Hence, the answer would be: 10.2 j