Answer:
Water normally freezes at 0°C (32°F). Salt lowers the freezing temperature. (That is, it can remain a liquid at much lower temperatures.)
When sprinkled on ice, the salt lowers the freezing temperature of the water which effectively melts the ice when the salt dissolves into it. There is a limit to how low it can reduce the temperature, though. If the temperature drops below -9°C (15°F), it's too cold for the salt to dissolve into the ice.
When making ice cream, the salt lowers the temperature of the ice and water sufficiently enough to freeze the cream.
Answer:

Explanation:
<u>Vertical Launch Upwards</u>
In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.
(a)
To find the initial speed we solve for vo:



(b)
The maximum time or the time taken by the object to reach its highest point is calculated as follows:



Academic- affect on substances, how it works
Scientific- helping cancer, making weapons
Answer:
E_{k2}=2660 [J] kinetic energy.
Explanation:
The energy in the initial state i.e. when the rollercoaster is at the top is equal to the energy in the final state i.e. when it is at the bottom of the hill.
These states can be represented by means of the second equation.
![E_{k1}+E_{p1}=E_{k2}\\160 + 2500 = E_{k2}\\E_{k2}=2660 [J]](https://tex.z-dn.net/?f=E_%7Bk1%7D%2BE_%7Bp1%7D%3DE_%7Bk2%7D%5C%5C160%20%2B%202500%20%3D%20E_%7Bk2%7D%5C%5CE_%7Bk2%7D%3D2660%20%5BJ%5D)
Since the rollercoaster is located in the bottom of the hill where the potential energy level is zero, therefore there is only kinetic energy in the second state.
Explanation:
(a) You can solve this using kinematics or energy.
Using kinematics:
a = F/m = 90 N / 4 kg = 22.5 m/s²
v₀ = 0 m/s
Δx = 5 m
Find: v
v² = v₀² + 2aΔx
v² = (0 m/s)² + 2 (22.5 m/s²) (5 m)
v = 15 m/s
Using energy:
W = ΔKE
Fd = ½ mv²
(90 N) (5 m) = ½ (4 kg) v²
v = 15 m/s
(b) ΔU = mg Δh
ΔU = (4 kg) (9.8 m/s) (12 m sin 40° − 15 m)
ΔU = -290 J
W = ΔU = -290 J