Here's my guesses.
1 sea stack ... A2 inlet ... E3 sand spit ... F4 sea arch ... B (looks like a land arch made by the sea ... ?)5 lagoon ... C6 barrier island ...A7 bay ... D8 sea cliff .... J9 sea cave .... G10 tombolo .... no idea
That's a lot of work for 9 points !!!!
Answer:
The shortest distance is
Explanation:
The free body diagram of this question is shown on the first uploaded image
From the question we are told that
The speed of the bicycle is ![v_b = 22\ mph = 22 * \frac{5280}{3600} = 32.26 ft/s](https://tex.z-dn.net/?f=v_b%20%3D%2022%5C%20mph%20%3D%2022%20%2A%20%5Cfrac%7B5280%7D%7B3600%7D%20%20%20%3D%20%2032.26%20ft%2Fs)
The distance between the axial is ![d = 42 \ in](https://tex.z-dn.net/?f=d%20%3D%2042%20%5C%20in)
The mass center of the cyclist and the bicycle is
behind the front axle
The mass center of the cyclist and the bicycle is
above the ground
For the bicycle not to be thrown over the
Momentum about the back wheel must be zero so
![\sum _B = 0](https://tex.z-dn.net/?f=%5Csum%20_B%20%3D%200)
=> ![mg (26) = ma(40)](https://tex.z-dn.net/?f=mg%20%2826%29%20%3D%20ma%2840%29)
=> ![a = \frac{26}{40} g](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B26%7D%7B40%7D%20g)
Here ![g = 32.2 ft/s^2](https://tex.z-dn.net/?f=g%20%3D%2032.2%20ft%2Fs%5E2)
So ![a = \frac{26}{40} (32.2)](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B26%7D%7B40%7D%20%2832.2%29)
Apply the equation of motion to this motion we have
![v^2 = u^2 + 2as](https://tex.z-dn.net/?f=v%5E2%20%3D%20u%5E2%20%2B%202as)
Where ![u = 32.26 ft /s](https://tex.z-dn.net/?f=u%20%3D%2032.26%20ft%20%2Fs)
and
since the bicycle is coming to a stop
![v^2 = (32.26)^2 - 2(20.93) S](https://tex.z-dn.net/?f=v%5E2%20%3D%20%2832.26%29%5E2%20-%202%2820.93%29%20S)
=>
Answer:
temperature
Explanation:
In general, the specific heat also depends on the temperature. The table below lists representative values of specific heat for various substances. Except for gases, the temperature and volume dependence of the specific heat of most substances is weak.
Answer:
The uncertainty in momentum changes by a factor of 1/2.
Explanation:
By Heisenberg's uncertainty principle, ΔpΔx ≥ h/2π where Δp = uncertainty in momentum and Δx = uncertainty in position = 0.2 nm. The uncertainty in momentum is thus Δp ≥ h/2πΔx. If the uncertainty in position is doubled, that is Δx₁ = 2Δx = 0.4 nm, the uncertainty in momentum Δp₁ now becomes Δp₁ ≥ h/2πΔx₁ = h/2π(2Δx) = (h/2πΔx)/2 = Δp/2.
So, the uncertainty in momentum changes by a factor of 1/2.
Answer:
![\huge\boxed{\sf F = 0.28\ N}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7B%5Csf%20F%20%3D%200.28%5C%20N%7D)
Explanation:
<h3>Given Data:</h3>
Mass = m = 68 kg
Velocity = v = 30 m/s
Time = 2 hours = 2 × 60 × 60 = 7200 s
<h3>Required:</h3>
Force = F = ?
<h3>Formula to be used:</h3>
![\displaystyle F = \frac{mv}{t}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%20%3D%20%5Cfrac%7Bmv%7D%7Bt%7D)
<h3>Solution:</h3>
![\displaystyle F = \frac{(68)(30)}{7200} \\\\F = \frac{2040}{7200} \\\\F = 0.28 N\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%20%3D%20%5Cfrac%7B%2868%29%2830%29%7D%7B7200%7D%20%5C%5C%5C%5CF%20%3D%20%5Cfrac%7B2040%7D%7B7200%7D%20%5C%5C%5C%5CF%20%3D%200.28%20N%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)