Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
#3, they both come after infared and are not harmful to living cells.
The density of the object is the ratio of its mass and volume. From the given dimensions above, we determine the volume through the equation,
V = L x W x H
Substituting,
V = (3 cm)(2 cm)(1 cm) = 6 cm³
From the idea presented above,
d = m/V
Substituting the known values,
d = (30 g)/ (6 cm³) = 5 g/cm³
ANSWER: 5 g/cm³
That is because it is impossible to create a law for the behavior of every single different gas, so creating laws for an ideal gas helps us understand the basic nature of gasses which might or might not differ slightly or a lot. By understanding how an ideal gas works, we can understand how a normal gas works.
