U = 0, initial vertical velocity
Neglect air resistance, and g = 9.8 m/s².
The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s
Answer: 2.0 s
<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2.
So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density.
So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave.
Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>
Let Karen's forward speed be considered as positive.
Therefore, before the headband is tossed backward, the speed of the headband is
V = 9 m/s
The headband is tossed backward relative to Karen at a speed of 20 m/s. Therefore the speed of the headband relative to Karen is
U = -20 m/s
The absolute speed of the headband, relative to a stationary observer is
V - U
= 9 + (-20)
= - 11 m/s
Answer:
The stationary observes the headband traveling (in the opposite direction to Karen) at a speed of 11 m/s backward.
Answer:
Explanation:
first write the newtons second law:
F
=δma
Applying bernoulli,s equation as follows:
∑
Where, 
is the pressure change across the streamline and 
is the fluid particle velocity
substitute 
for {tex]γ[/tex] and 
for 
integrating the above equation using limits 1 and 2.
there the bernoulli equation for this flow is
note: 
=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular
Answer:
x = v₀ cos θ t
, y = y₀ + v₀ sin θ t - ½ g t2
Explanation:
This is a projectile launch exercise, in this case we will write the equations for the x and y axes
Let's use trigonometry to find the components of the initial velocity
sin θ = 
/ v₀
cos θ = v₀ₓ / v₀
v_{y} = v_{oy} sin θ
v₀ₓ = vo cos θ
now let's write the equations of motion
X axis
x = v₀ₓ t
x = v₀ cos θ t
vₓ = v₀ cos θ
Y axis
y = y₀ + t - ½ g t2
y = y₀ + v₀ sin θ t - ½ g t2
v_{y} = v₀ - g t
v_{y} = v₀ sin θ - gt
= v_{oy}^2 sin² θ - 2 g y
As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components
