Answer: Gravitational force and drag force
Explanation:
For a snowboard jumper in the air, two forces would be acting. One in the downward direction- the gravitational pull and second in the opposite direction to the motion, the drag force due to air. If the snowboard jumper jumps in the air at a certain angle with the horizontal. The forces are written as the sum of vertical and horizontal components. Hence, for the modeling the motion, gravitational force and drag force are important,
Answer: You could dissolve it by heating it back up, then just cooling it down again.
Hope that helps!
Answer:
680 J
Explanation:
Mechanical energy = potential energy + kinetic energy
ME = PE + KE
ME = mgh + ½ mv²
ME = (77.1 kg) (9.8 m/s²) (0.90 m) + ½ (77.1 kg) (0 m/s)²
ME = 680 J
Answer: The bottom of the ladder is moving at 3.464ft/sec
Explanation:
The question defines a right angle triangle. Therefore using pythagorean
h^2 + l^2 = 10^2 = 100 ...eq1
dh/dt = -2ft/sec
dl/ dt = ?
Taking derivatives of time in eq 1 on both sides
2hdh/dt + 2ldl/dt = 0 ....eq2
Putting l = 5ft in eq2
h^ + 5^2 = 100
h^2 = 25 = 100
h Sqrt(75)
h = 8.66 ft
Put h = 8.66ft in eq2
2 × 8.66 × (-2) + 2 ×5 dl/dt
dl/dt = 17.32 / 5
dl/dt = 3.464ft/sec
Answer:
w = 4,786 rad / s
, f = 0.76176 Hz
Explanation:
For this problem let's use the concept of angular momentum
L = I w
The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved
Initial Before sticking
L₀ = 0 + I₂ w₂
Final after coupling
= (I₁ + I₂) w
The moments of inertia of a disk with an axis of rotation in its center are
I = ½ M R²
How the moment is preserved
L₀ = 
I₂ w₂ = (I₁ + I₂) w
w = w₂ I₂ / (I₁ + I₂)
Let's reduce the units to the SI System
d₁ = 60 cm = 0.60 m
d₂ = 40 cm = 0.40 m
f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz
Angular velocity and frequency are related.
w₂ = 2 π f₂
w₂ = 2π 3.33
w₂ = 20.94 rad / s
Let's replace
w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)
w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)
Let's calculate
w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)
w = 20.94 1.28 / 5.6
w = 4,786 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 4.786 / 2π
f = 0.76176 Hz