Answer:
11:1
Explanation:
At constant acceleration, an object's position is:
y = y₀ + v₀ t + ½ at²
Given y₀ = 0, v₀ = u, and a = -g:
y = u t − ½g t²
After 6 seconds, the ball reaches the maximum height (v = 0).
v = at + v₀
0 = (-g)(6) + u
u = 6g
Substituting:
y = 6g t − ½g t²
The displacement between t=0 and t=1 is:
Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]
Δy = 6g − ½g
Δy = 5½g
The displacement between t=6 and t=7 is:
Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]
Δy = (42g − 24½g) − (36g − 18g)
Δy = 17½g − 18g
Δy = -½g
So the ratio of the distances traveled is:
(5½g) / (½g)
11 / 1
The ratio is 11:1.
Answer:
b) total energy input equals total energy output
Explanation:
The first law of thermodynamics is a generalization of the conservation of energy in thermal processes. It is based on Joule's conclusion that heat and energy are equivalent. But to get there you have to get around some traps along the way.
From Joule's conclusion we might be tempted to call heat "internal" energy associated with temperature. We could then add heat to the potential and kinetic energies of a system, and call this sum the total energy, which is what it would conserve. In fact, this solution works well for a wide variety of phenomena, including Joule's experiments. Problems arise with the idea of heat "content" of a system. For example, when a solid is heated to its melting point, an additional "heat input" causes the melting but without increasing the temperature. With this simple experiment we see that simply considering the thermal energy measured only by a temperature increase as part of the total energy of a system will not give a complete general law.
Instead of "heat," we can use the concept of internal energy, that is, an energy in the system that can take forms not directly related to temperature. We can then use the word "heat" to refer only to a transfer of energy between a system and its environment. Similarly, the term work will not be used to describe something contained in the system, but describes a transfer of energy from one system to another. Heat and work are, therefore, two ways in which energy is transferred, not energies.
In an isolated system, that is, a system that does not exchange matter or energy with its surroundings, the total energy must remain constant. If the system exchanges energy with its environment but not matter (what is called a closed system), it can do so only in two ways: a transfer of energy either in the form of work done on or by the system, either in the form of heat to or from the system. In the event that there is energy transfer, the change in the energy of the system must be equal to the net energy gained or lost by the environment.
Answer:
= 5.1 W
Explanation:
time (t) = 30 ms = 0.03 s
mass (m) = 560 g = 0.56 kg
initial velocity (U) = 0 m/s
final velocity (V) = 0.74 m/s
power = \frac{work done}{t} = \frac{f x d}{t} = f x v = m x a x v
m x a x v = m x \frac{V-U}{t} x \frac{V + U}{2}
m x \frac{V-U}{t} x \frac{V + U}{2} = 0.56 x \frac{0.74 - 0}{0.03} x \frac{0.74+0}{2}
= 5.1 W
Answer:
45 m
Explanation:
Use Big Five #5 with v0 = 0 (calling down the positive direction).
