2. A 3.0 kg block rests on a level surface. The coefficient of static friction is µs = 0.50, and the coefficient of kinetic fric
1 answer:
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Answer:
326149.2 KJ
Explanation:
The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:
Q = m*cv*ΔT
Where m is the mass of the object, cv is the specific heat capacity at constant volume, which basically means the amount of heat necessary for a 1kg of water to increase 1C degree in temperatur, and ΔT is the change in temperature.
A 65000 L swimming pool will have a mass of:
65000L * = 65000 kg
The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.
We replace the data and get:
Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ
Answer:
A) t = 0.55 s
B) x = 24.8 m
Explanation:
A) We can find the time at which the ball will be in the air using the following equation:
Where:
is the final height= 0
is the initial height= 1.5 m
is the component of the initial speed in the vertical direction = 0 m/s
t: is the time =?
g: is the gravity = 9.81 m/s²
By solving the above equation for t we have:
Hence, the ball will stay 0.55 seconds in the air.
B) We can find the distance traveled by the ball as follows:
Where:
a: is the acceleration in the horizontal direction = 0
is the final position =?
is the initial position = 0
is the component of the initial speed in the horizontal direction = 45 m/s
Therefore, the ball will travel 24.8 meters.
I hope it helps you!