Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
Explanation:
∆x=300 m×2
∆t=1.5 s
v=∆x/∆t → v=2×300/1.5 = 400 m/s
Answer: The molar heat capacity of aluminum is 
Explanation:
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of water = 130.0 g
= mass of aluminiunm = 23.5 g
= final temperature
= 
= temperature of water = 
= temperature of aluminium = 
= specific heat of water= 
= specific heat of aluminium= ?
Now put all the given values in equation (1), we get
![130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]](https://tex.z-dn.net/?f=130.0%5Ctimes%204.184%5Ctimes%20%28299-296%29%3D-%5B23.5%5Ctimes%20c_2%5Ctimes%20%28299-373%29%5D)
Molar mass of Aluminium = 27 g/mol
Thus molar heat capacity =
