When you breathe in, or inhale, your diaphragm contracts (tightens) and moves downward. This increases the space in your chest cavity, into which your lungs expand. The intercostal muscles between your ribs also help enlarge the chest cavity. They contract to pull your rib cage both upward and outward when you inhale.
Answer:
5.25 m
Explanation:
Given;
The height equation h;
h=-x^2+3x+3
Where;
h = the height above water
x = horizontal distance from the end of the board
The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.
differentiating the equation h.
dh/dx = h' = -2x + 3 = 0
Solving for x;
2x = 3
x = 3/2
Substituting into the function h;
h max = -x^2+3x+3
h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =
h max = 21/4 = 5.25 m
Explanation:
(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.
Sum of forces in the centripetal direction:
∑F = ma
mg = mv²/RL
v = √(g RL)
(b) Energy is conserved.
EE = KE + RE + PE
½ kd² = ½ mv² + ½ Iω² + mgh
kd² = mv² + Iω² + 2mgh
kd² = mv² + (m RC²) ω² + 2mg (2 RL)
kd² = mv² + m RC²ω² + 4mg RL
kd² = mv² + mv² + 4mg RL
kd² = 2mv² + 4mg RL
kd² = 2m (v² + 2g RL)
d² = 2m (v² + 2g RL) / k
d = √[2m (v² + 2g RL) / k]
Answer:
v₂ = 176.24 m/s
Explanation:
given,
angle of projectile = 45°
speed = v₁ = 150 m/s
for second trail
speed = v₂ = ?
angle of projectile = 37°
maximum height attained formula,
now,
now, equating both the equations
v₂² = 31061.79
v₂ = 176.24 m/s
velocity of projectile would be equal to v₂ = 176.24 m/s
Answer:
B. over the symbol.
Explanation:
vectors are represented with a symbol carrying an arrow head with also indicates direction
