Answer:
Explanation:
Given
Airplane is flying with horizontally with a constant momentum during time interval 
Impulse is given by change in momentum, so there is no net impulse on the Plane because momentum is constant
(b) As there is no change in momentum therefore impulse of thrust and air drag is balanced i.e. both are equal in magnitude but act in opposite direction
Arrhenius' equation relates the dependence of rate constant of a chemical reaction to the temperature. The equation below is the Arrhenius equation
where k is the rate constant, T is the absolute temperature. As the temperature of the system increases, the rate constant also increases and vice versa.
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
Answer:
the can's kinetic energy is 0.42 J
Explanation:
given information:
Mass, m = 460 g = 0.46 kg
diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m
velocity, v = 1.1 m/s
the kinetic energy of the can is the total of kinetic energy of the translation and rotational.
KE = 
I ω^2 + 
where
I = 
and ω = 
thus,
KE = ()^2 +
= 
+
= 
+
= 
= 
= 0.42 J
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