Final answer:
The kinetic energy of the Sherman M4 medium tank traveling at its top speed can be calculated using the formula KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
The kinetic energy of the Sherman M4 medium tank traveling at its top speed of 39 km/h is approximately 1.78 Megajoules (MJ).
Step-by-step explanation:
To calculate the kinetic energy of a Sherman M4 medium tank traveling at its top speed, we use the kinetic energy formula:
KE = \( \frac{1}{2}mv^2 \)
where:
- KE is the kinetic energy,
- m is the mass of the tank, and
- v is the velocity of the tank.
Given that the mass m of the Sherman tank is 30,300 kg and its top speed v is 39 km/h (which needs to be converted to meters per second by multiplying by \( \frac{1000}{3600} \)), we find:
v = 39 km/h = 39 \( \frac{1000}{3600} \) m/s = 10.833 m/s
Now we can plug these values into the kinetic energy formula:
KE = \( \frac{1}{2} \) \( 30300 \) kg \( (10.833 m/s)^2 \)
KE = \( \frac{1}{2} \) \( 30300 \) kg \( 117.317 \) m2/s2
KE = \( 30300 \) kg \( 58.6585 \) m2/s2
KE = 1,777,411.95 J or approximately 1.78 MJ
Therefore, the kinetic energy of the Sherman M4 medium tank traveling at its top speed of 39 km/h is approximately 1.78 Megajoules (MJ).