Answer:
10 m/s
Explanation:
The potential energy of the roller coaster is converted to kinetic energy. According to law of transformation of energy, potential energy = kinetic energy.
mgh = ½mv².Take g as 10m/s².
150 * 10 * 50 = ½ * 150 * v².
7500 = 75v².
v² = 7500/75.
v = √100.
v= 10 m/s.
Hi there!
We can use the following equation for constant velocity:

d = displacement (m)
v = velocity (m/s)
t = time (s)
Plug in the givens:

Answer:

Explanation:
We have to many data in different system, so we need transform everything to SI, that is

When we have all this values in SI apply a Energy Balance Equation,
![\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0](https://tex.z-dn.net/?f=%5Cdot%7BQ%7D_%7Bcv%7D-%5Cdot%7BW%7D_%7Bcv%7D%2B%5Cdot%7Bm%7D%5B%28h_1-h_2%29%2B%28%5Cfrac%7BV_1%5E2-V_2%5E2%7D%7B2%7D%29%2Bg%28z_1-z_2%29%5D%3D0)
Solving for V_2

From the table of gas properties we calculate for
and 


For T_2;

Substituting in equation for V_2


Answer:

Explanation:
First we need to state our assumptions:
Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice 
Mass of water,
.
Energy balance for the ice-water system is defined as

a.The mass of ice at
is defined as:
![[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g](https://tex.z-dn.net/?f=%5Bmc%280-T_1%29%2Bmh_i_f%2Bmc%28T_2-0%29%5D_i_c_e%2B%5Bmc%28T_2-T_1%29%5D_w%3D0%5C%5C%5C%5Cm_i_c_e%5B0%2B333.7%2B418%5Ctimes%285-0%29%5D%2B0.3%5Ctimes4.18%5Ctimes%285-20%29%3D0%5C%5Cm_i_c_e%3D0.0546Kg%3D54.6g)
b.Mass of ice at
is defined as:
![[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g](https://tex.z-dn.net/?f=%5Bmc%280-T_1%29%2Bmh_i_f%2Bmc%28T_2-0%29%5D_i_c_e%2B%5Bmc%28T_2-T_1%29%5D_w%3D0%5C%5C%5C%5Cm_i_c_e%5B2.11%5Ctimes%280-%2820%29%29%2B333.7%2B4.18%5Ctimes%285-0%29%5D%2B0.3%5Ctimes4.18%5Ctimes%285-20%29%3D0%5C%5C%5C%5Cm_i_c_e%3D0.0487Kg%3D48.7g)
c.Mass of cooled water at 

![[mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g](https://tex.z-dn.net/?f=%5Bmc%28T_2-T_1%29%5D_c_w%2B%5Bmc%28T_2-T_1%29%5D_w%3D0%5C%5Cm_c_w%5Ctimes4.18%5Ctimes%285-0%29%2B0.3%5Ctimes4.18%5Ctimes%285-20%29%5C%5Cm_c_w%3D0.9kg%3D900g)