Answer:
The terminal velocity is ![v_t =17.5 \ m/s](https://tex.z-dn.net/?f=v_t%20%20%3D17.5%20%5C%20m%2Fs)
Explanation:
From the question we are told that
The mass of the squirrel is ![m_s = 50\ g = \frac{50}{1000} = 0.05 \ kg](https://tex.z-dn.net/?f=m_s%20%20%3D%20%2050%5C%20g%20%20%3D%20%20%5Cfrac%7B50%7D%7B1000%7D%20%3D%20%200.05%20%5C%20%20kg)
The surface area is ![A_s = 935 cm^2 = \frac{935}{10000} = 0.0935 \ m^2](https://tex.z-dn.net/?f=A_s%20%3D%20%20935%20cm%5E2%20%20%3D%20%20%5Cfrac%7B935%7D%7B10000%7D%20%3D%200.0935%20%5C%20m%5E2)
The height of fall is h =4.8 m
The length of the prism is ![l = 23.2 = 0.232 \ m](https://tex.z-dn.net/?f=l%20%3D%20%2023.2%20%3D%200.232%20%5C%20m)
The width of the prism is ![w = 11.6 = 0.116 \ m](https://tex.z-dn.net/?f=w%20%3D%20%2011.6%20%3D%20%200.116%20%5C%20m)
The terminal velocity is mathematically represented as
![v_t = \sqrt{\frac{2 * m_s * g }{\dho_s * C * A } }](https://tex.z-dn.net/?f=v_t%20%20%3D%20%20%5Csqrt%7B%5Cfrac%7B2%20%2A%20m_s%20%2A%20%20g%20%7D%7B%5Cdho_s%20%2A%20C%20%20%2A%20A%20%7D%20%7D)
Where
is the density of a rectangular prism with a constant values of ![\rho = 1.21 \ kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%20%3D%20%201.21%20%5C%20kg%2Fm%5E3)
is the drag coefficient for a horizontal skydiver with a value = 1
A is the area of the prism the squirrel is assumed to be which is mathematically represented as
![A = 0.116 * 0.232](https://tex.z-dn.net/?f=A%20%3D%20%200.116%20%2A%200.232)
![A = 0.026912 \ m^2](https://tex.z-dn.net/?f=A%20%3D%20%200.026912%20%5C%20m%5E2)
substituting values
![v_t = \sqrt{\frac{2 * 0.510 * 9.8 }{1.21 * 1 * 0.026912 } }](https://tex.z-dn.net/?f=v_t%20%20%3D%20%20%5Csqrt%7B%5Cfrac%7B2%20%2A%200.510%20%2A%20%209.8%20%7D%7B1.21%20%2A%201%20%20%2A%200.026912%20%7D%20%7D)
![v_t =17.5 \ m/s](https://tex.z-dn.net/?f=v_t%20%20%3D17.5%20%5C%20m%2Fs)
Answer:
1. the pencil would have the momentum and would keep going until it hits the windshield. 2. when the car suddenly accelerates, the pencil would be inert and it would move toward the back of the car until a constant speed from the car is reached.
Answer:
The increase in temperature of the bullet is 351.1 kelvin
Explanation:
First, we should find the kinetic energy of the bullet is:
![K=\frac{mv^{2}}{2}=\frac{(6\times10^{-3}\,kg)(420\,\frac{m}{s})^{2}}{2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7Bmv%5E%7B2%7D%7D%7B2%7D%3D%5Cfrac%7B%286%5Ctimes10%5E%7B-3%7D%5C%2Ckg%29%28420%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%29%5E%7B2%7D%7D%7B2%7D%20)
with m the mass and v the velocity.
![K=529.2 J](https://tex.z-dn.net/?f=K%3D529.2%20J%20)
Now we know that half of the kinetic energy of the bullet is transformed into internal energy, by second's law of thermodynamics that means heat (Q) to raise bullet temperature (T), so:
![Q=\frac{K}{2}= 264.6\,J](https://tex.z-dn.net/?f=%20Q%3D%5Cfrac%7BK%7D%7B2%7D%3D%20264.6%5C%2CJ)
To know what the increase in temperature is, we should use specific heat of lead:
![c=125.604 \frac{J}{kg\.K}](https://tex.z-dn.net/?f=c%3D125.604%20%5Cfrac%7BJ%7D%7Bkg%5C.K%7D%20)
The equation that relates specific heat, change in temperature and mass is:
![Q=cm\varDelta T](https://tex.z-dn.net/?f=%20Q%3Dcm%5CvarDelta%20T)
solving for
:
![\varDelta T=\frac{Q}{cm}=\frac{264.6}{(125.604)(6\times10^{-3})}](https://tex.z-dn.net/?f=%5CvarDelta%20T%3D%5Cfrac%7BQ%7D%7Bcm%7D%3D%5Cfrac%7B264.6%7D%7B%28125.604%29%286%5Ctimes10%5E%7B-3%7D%29%7D%20)
![\varDelta T= 351.10\, K](https://tex.z-dn.net/?f=%20%5CvarDelta%20T%3D%20351.10%5C%2C%20K)