If a 1500 kg car is moving 20.0 m/s is brought to a stop by brakes. How many kilocalories of heat are generated in the process?
Please include the formula that I should use to find the answer.
-Thanks
72 kcal
Wouldn't the Kinetic energy all turn to heat? KE= 1/2 m v^2
YOu will have to change it to Kcal, as KE is in joules. You can check your calcs by pasting this into Google search window
1/2 * 1500kg * (20m/s)^2 in kilocalories
I tried pasting the equation into google but,nothing happened
A 1500 kg car moving at 20.0 m/s is brought to a stop by the brakes. How many kilocalories of heat are generated in this process?
To find the amount of heat generated when a car is brought to a stop, we can use the following formula:
Q = m * v^2 / (2 * R)
Where:
Q is the heat generated (in kilocalories)
m is the mass of the car (in kilograms)
v is the velocity of the car (in meters per second)
R is the deceleration rate (in meters per second squared)
In this case, the car's mass (m) is given as 1500 kg and the velocity (v) is given as 20.0 m/s. To find the deceleration rate (R), we need to know the distance over which the car comes to a stop. However, this information is not provided in the question.
Therefore, without knowing the distance over which the car is brought to a stop, we cannot calculate the exact amount of heat generated using the given information.
If you have the distance information, you can use the formula Q = m * v^2 / (2 * R) to find the answer.