An object with more mass has more kinetic energy than an object with less mass, if both objects are moving at the same speed. <em>(c)</em>
Answer:
1) 3.92 J
2) 1596.08 J
3) 16.3 s ??
Explanation:
Initial Potential energy PE = mgh = 0.5(9.8)(0) = 0 J
Initial Kinetic energy KE = ½mv² = ½(0.5)80² = 1600 J
PE = 0.5(9.8)(0.80) = 3.92 J
KE = 1600 - 3.92 = 1596.08 J
Question 3 is not clear
to the point 80 cm above the ground the flight time is only 0.01 s
The time when the mass strikes ground again will be twice the time gravity takes to reduce the initial velocity to zero
t = 2(80.0 / 9.8) = 16.3 s
would not 80 m above the ground be a much more interesting point to consider?
PE = 0.5(9.8)(80) = 392 J
KE = 1600 - 392 = 1208 J
v₈₀ = √(2(1280) /0.5) = 69.5 m/s
t₈₀ = h/v(avg) = 80 / (½(80 + 69.5)) = 1.07 s
Answer:
final equilibrium temperature is 65.02 ◦C
Explanation:
Given data
milk = 12 g
temp milk = 7◦C
coffee = 175 g
coffee temp = 69◦C
to find out
final equilibrium temperature
solution
we apply here equilibrium condition i.e m c ΔT
milk mass × c × ΔT = coffee mass × c × ΔT
12 × 1 × (T- 7) = 175 × 1 × ( 69- T)
12T - 84 = 12075 - 175 T
187 T = 12159
T = 65.02
so final equilibrium temperature is 65.02 ◦C
Can't really plot a graph here for question 1.
2a) The car speeds up from A to B. The car travels at a constant speed from B to C. The car slows down to a stop from C to D.
b) From the graph, at 10 seconds, the car is moving at 20 m/s.
The block on the action of two forces, the Force of Friction and the Tangential Weight. Using the Newton's Secound Law, we have:
Using the Velocity Hourly Equation, we get:
Uniting the equations:
Entering the unknowns:
Obs: Approximate results
If you notice any mistake in my english, please let me know, because i am not native.
