TRUE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Given:
Amount of heat produced = 100 kcal per hour
Let's find the rate of energy production in joules.
We know that:
1 calorie = 4.184 Joules
1 kcal = 4.184 Joules
To find the rate of energy production in Joules, we have:
Therefore, the rate of energy production in joules is 418.4 kJ/h which is equivalent to 418400 Joules
ANSWER:
418.4 kJ/h
Answer:
Car H
Explanation:
Frictional force is a resistant force. It is given as:
F = u*m*g
Where u = coefficient of friction
m = mass
g = acceleration due to gravity
From the formula above, we see that frictional force is dependent on the mass of object and the coefficient of friction.
Since they all have the same tires, the coefficient of friction between the tire and the floor is the same for each car. Acceleration due to gravity, g, is constant.
The only factor that determines the frictional force of each car is the mass. Hence, the more the mass, the more the frictional force.
So, the most massive car will have the most frictional force and hence, will come to a stop quicker than the others. The least massive car will have the least frictional force and so, will take a longer time to stop.
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
