Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force =
Equating both, we have
So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
Answer:
There is 148.35 Joules of heat is released in the process.
Explanation:
Given that,
Heat capacity of the object,
Initial temperature,
Final temperature,
We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :
Let the mass of the object is 10 g or 0.01 kg
So,
Q = 148.35 Joules
So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m