Answer:

Explanation:
We can use the following SUVAT equation to solve the problem:

where
v = 0 is the final velocity of the car
u = 24 m/s is the initial velocity
a is the acceleration
d = 196 m is the displacement of the car before coming to a stop
Solving the equation for a, we find the acceleration:

Answer:
73.5 m/s
Explanation:
The position of the first ball is:
y = y₀ + v₀ t + ½ at²
y = h + (0)(18) + ½ (-9.8)(18)²
y = h − 1587.6
The position of the second ball is:
y = y₀ + v₀ t + ½ at²
y = h + (-v) (18−6) + ½ (-9.8)(18−6)²
y = h − 12v − 705.6
Setting the positions equal:
h − 1587.6 = h − 12v − 705.6
-1587.6 = -12v − 705.6
1587.6 = 12v + 705.6
882 = 12v
v = 73.5
The second ball is thrown downwards with a speed of 73.5 m/s
Answer:
When adolescents get enough calcium during the teen years, they can start out their adult lives with the strong bones and significantly reduce their risk for fractures as an adult. Inadequate calcium intake during adolescence and young adulthood puts individuals at risk for developing osteoporosis later in life.
Answer:6m/s^2
Explanation:
Mass =3kg
Force=18N
Acceleration =force ➗ mass
Acceleration =18 ➗ 3
Acceleration =6
Acceleration =6m/s^2
Answer:
The final equilibrium T_{f} = 25.7[°C]
Explanation:
In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.
That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.
The temperature of thermal equilibrium will be T_f.
The heat absorbed by water will be equal to the heat rejected by Iron.

Heat transfer can be found by means of the following equation.

where:
Qiron = Iron heat transfer [kJ]
m = iron mass = 200 [g] = 0.2 [kg]
T_i = Initial temperature of the iron = 300 [°C]
T_f = final temperature [°C]

Cp_iron = 437 [J/kg*°C]
Cp_water = 4200 [J/kg*°C]
![0.2*437*(300-T_{f})=1*4200*(T_{f}-20)\\26220-87.4*T_{f}=4200*T_{f}-84000\\26220+84000=4200*T_{f}+87.4*T_{f}\\110220 = 4287.4*T_{f}\\T_{f}=25.7[C]](https://tex.z-dn.net/?f=0.2%2A437%2A%28300-T_%7Bf%7D%29%3D1%2A4200%2A%28T_%7Bf%7D-20%29%5C%5C26220-87.4%2AT_%7Bf%7D%3D4200%2AT_%7Bf%7D-84000%5C%5C26220%2B84000%3D4200%2AT_%7Bf%7D%2B87.4%2AT_%7Bf%7D%5C%5C110220%20%3D%204287.4%2AT_%7Bf%7D%5C%5CT_%7Bf%7D%3D25.7%5BC%5D)