Work package. Hope this helps!
Answer:
a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m
Explanation:
Mass of the dart = 0.02kg, the spring was compressed to 6cm
Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m
Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m
Work needed to compress the spring = 0.036J
b) the total energy stored in the spring = the work done to compress the spring = 0.036J
c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.
d) 1/2mv^2 = 0.036
mv^2 = 0.036*2
v^2 = 0.036*2 / 0.02 = 3.6
v = √3.6 = 1.897 approx 1.9m/s
e) kinetic energy of the dart = work done against gravity to get the body to height h
Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward
0.036 = 0.02 * 9.81 * h
0.036 / ( 0.02*9.81) = h
h = 0.18 m
<span>The initial speed, u of plane in terms of velocity of sound which may be taken as U
u=142/331=0.429*U
It crosses the sound barrier after says t seconds then we have 331-142=23.1*t or t is given 8.18 s exactly t=9/11s.
After 18 seconds the plane will traveling with velocity V
V=142+18*23.1=557.8 m/s==1.685*U</span>
Answer : The specific heat of unknown sample is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
![q_1=-[q_2+q_3]](https://tex.z-dn.net/?f=q_1%3D-%5Bq_2%2Bq_3%5D)
![m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_f-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_f-T_2%29%2Bm_3%5Ctimes%20c_3%5Ctimes%20%28T_f-T_2%29%5D)
where,
= specific heat of unknown sample = ?
= specific heat of water = 
= specific heat of copper = 
= mass of unknown sample = 72.0 g = 0.072 kg
= mass of water = 203 g = 0.203 kg
= mass of copper = 187 g = 0.187 kg
= final temperature of calorimeter = 
= initial temperature of unknown sample = 
= initial temperature of water and copper = 
Now put all the given values in the above formula, we get
![0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]](https://tex.z-dn.net/?f=0.072kg%5Ctimes%20c_1%5Ctimes%20%2839.4-80.0%29%5EoC%3D-%5B%280.203kg%5Ctimes%204186J%2Fkg%5EoC%5Ctimes%20%2839.4-11.0%29%5EoC%29%2B%280.187kg%5Ctimes%20390J%2Fkg%5EoC%5Ctimes%20%2839.4-11.0%29%5EoC%29%5D)
Therefore, the specific heat of unknown sample is,
The boiling point is defined as the temperature at which the pressure of the vapor of the liquid is equivalent to the external atmospheric pressure surrounding the liquid. Therefore, the boiling point of the liquid is dependent on the atmospheric pressure.
Based on this, the vapor pressure of cyclohexane at 81 degrees celcius will be equal to atmospheric pressure (based on barometric readings)<span />
