Answer:
40.3 min
Explanation:
First of all, let's convert every quantity into SI units:
(speed in the first part of the trip)
time during which the person has stopped
(average speed of the whole trip)
The average speed is the ratio between the total distance covered, d, and the total time taken, t:
(1)
The total distance covered is simply
![d = v_1 t_1](https://tex.z-dn.net/?f=d%20%3D%20v_1%20t_1)
where
is the time during which the person has moved at 92.5 km/h.
The total time taken is
![t= t_1 + t_2](https://tex.z-dn.net/?f=t%3D%20t_1%20%2B%20t_2)
So (1) becomes
![v=\frac{v_1 t_1}{t_1 + t_2}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bv_1%20t_1%7D%7Bt_1%20%2B%20t_2%7D)
Solving for
:
![v t_1 + v t_2 = v_1 t_1\\vt_2 = (v_1+v)t_1\\t_1 = \frac{v t_2}{v_1+v}=\frac{(20.1 m/s)(1680 s)}{25.7 m/s + 20.1 m/s}=737.3 s](https://tex.z-dn.net/?f=v%20t_1%20%2B%20v%20t_2%20%3D%20v_1%20t_1%5C%5Cvt_2%20%3D%20%28v_1%2Bv%29t_1%5C%5Ct_1%20%3D%20%5Cfrac%7Bv%20t_2%7D%7Bv_1%2Bv%7D%3D%5Cfrac%7B%2820.1%20m%2Fs%29%281680%20s%29%7D%7B25.7%20m%2Fs%20%2B%2020.1%20m%2Fs%7D%3D737.3%20s)
which corresponds to
![t_2 = 737.3 s = 12.3 min](https://tex.z-dn.net/?f=t_2%20%3D%20737.3%20s%20%3D%2012.3%20min)
So the total time of the trip is
![t = 28.0 min + 12.3 min = 40.3 min](https://tex.z-dn.net/?f=t%20%3D%2028.0%20min%20%2B%2012.3%20min%20%3D%2040.3%20min)
Newton's second law is:
F=m*a,
where a=dv/dt, so
F=m*(dv/dt)
Rearranging gives:
F*dt=m*dv.
Basic integration gives:
F*t=m(vf-v0),
where vf and v0 are the final and initial velocities of the object respectively.
In your case vf=0, because the ball stops completely, and v0=10m/s.
Rearranging the last expression gives F=(m(vf-vo))/t.
Plug in numbers to find F=(2*10)/0.03=666.6 N
Answer:
The value is ![V_2 = 1.9396 *10^{-5} \ m^3](https://tex.z-dn.net/?f=V_2%20%3D%201.9396%20%2A10%5E%7B-5%7D%20%5C%20%20m%5E3%20)
Explanation:
From the question we are told that
The depth at which the bubble is released is ![h = 115 \ m](https://tex.z-dn.net/?f=h%20%3D%20115%20%5C%20%20m)
The volume of the air bubble is ![V = 1.60 cm^3 = 1.60 *10^{-6} \ m^3](https://tex.z-dn.net/?f=V%20%20%3D%20%201.60%20cm%5E3%20%3D%201.60%20%2A10%5E%7B-6%7D%20%5C%20%20m%5E3)
Generally from the ideal gas law
![PV = nRT](https://tex.z-dn.net/?f=PV%20%20%3D%20%20nRT)
Given that n , R , T are constant we have that
![PV = constant](https://tex.z-dn.net/?f=PV%20%20%3D%20%20constant)
So
![P_1 V_1=P_2 V_2](https://tex.z-dn.net/?f=P_1%20V_1%3DP_2%20V_2)
Here
is the pressure of the bubble at the depth where it is released which i mathematically represented as
![P_1 = P_a + P](https://tex.z-dn.net/?f=P_1%20%3D%20P_a%20%2B%20P)
Here
is the atmospheric pressure with value ![P_a = 101325 \ Pa](https://tex.z-dn.net/?f=P_a%20%20%3D%20%20101325%20%5C%20%20Pa)
and
is the pressure due to the depth which is mathematically represented as
![P = \rho * g * h](https://tex.z-dn.net/?f=P%20%3D%20%5Crho%20%2A%20%20g%20%2A%20h)
So
![P = 1000 * 9.8*115](https://tex.z-dn.net/?f=P%20%3D%201000%20%2A%209.8%2A115)
=> ![P = 1127000\ Pa](https://tex.z-dn.net/?f=P%20%3D%201127000%5C%20Pa)
Here
is the density of pure water with value ![\rho = 1000 \ kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%201000%20%5C%20kg%2Fm%5E3)
![g = 9.8 \ m/s^2](https://tex.z-dn.net/?f=g%20%3D%209.8%20%5C%20%20m%2Fs%5E2)
is the volume of the bubble at the depth where it is released
is the pressure of the bubble at the surface which is equivalent to the atmospheric temperature
is the volume of the bubble at the surface
So
![V_2 = \frac{ P_1 * V_1}{ P_2}](https://tex.z-dn.net/?f=V_2%20%3D%20%5Cfrac%7B%20P_1%20%2A%20V_1%7D%7B%20P_2%7D)
=> ![V_2 = \frac{(Pa+ P) * V_1}{P_a}](https://tex.z-dn.net/?f=V_2%20%3D%20%5Cfrac%7B%28Pa%2B%20P%29%20%2A%20V_1%7D%7BP_a%7D)
=> ![V_2 = \frac{101325 + 1127000 * (1.60 *10^{-6})}{ 101325 }](https://tex.z-dn.net/?f=V_2%20%3D%20%5Cfrac%7B101325%20%2B%201127000%20%20%20%2A%20%281.60%20%2A10%5E%7B-6%7D%29%7D%7B%20101325%20%7D)
=> ![V_2 = 1.9396 *10^{-5} \ m^3](https://tex.z-dn.net/?f=V_2%20%3D%201.9396%20%2A10%5E%7B-5%7D%20%5C%20%20m%5E3%20)