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.
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.
First, we determine how long the parcel will fall using:
s = ut + 1/2 at²
where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.
5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds
Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time
The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.
Distance = 10 * 1.06
Distance = 10.6 meters
The boat should be 10.6 meters away horizontally from the point of release.
Answer:
a. 8.33 x 10 ⁻⁶ Pa
b. 8.19 x 10 ⁻¹¹ atm
c. 1.65 x 10 ⁻¹⁰ atm
d. 2.778 x 10 ⁻¹⁴ kg / m²
Explanation:
Given:
a.
I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s
P rad = I / us
P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s
P rad = 8.33 x 10 ⁻⁶ Pa
b.
P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]
P rad = 8.19 x 10 ⁻¹¹ atm
c.
P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]
P rad = 1.67 x 10 ⁻⁵ Pa
P₁ = 1.013 x 10 ⁵ Pa /atm
P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm
d.
P rad = I / us
ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²
ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²
Total time in between the dropping of the stone and hearing of the echo = 8.9 s
Time taken by the sound to reach the person = 0.9 s
Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds
Initial speed (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s^2
Time taken (t) = 8 seconds
Let the depth of the well be h.
Using the second equation of motion:

h = 313.6 m
Hence, the depth of the well is 313.6 m
Very roughly 7,700 feet ... about 1.5 miles.