Answer: The final temperature is 470K
Explanation: Using the relation;
Q= ΔU +W
Given, n = 2mol
Initial temperature T1= 345K
Heat =Q= 2250J
Workdone=W=-870J(work is done on gas)
T2 =Final temperature =?
ΔU =3/2nR(T2-T1)
ΔU=3/2 × 2 ×8.314 (T2 - 345)
ΔU=24.942(T2-345)
Therefore Q = 24.942(T2-345)+ (-870)
2250=24.942(T2-345)+ (-870)
125.09=(T2-345)
T2 =470K
Therfore the final temperature is 470K
<span>A+B-C
</span><span>A = 6x - 2y
B = -4x - 8y
C = -3x + 9y
(</span>6x - 2y) + (-4x - 8y) - (-3x + 9y)
(6x - 2y) + (-4x - 8y) + (3x - 9y)
2x -10y + (3x - 9y)
5x - 19y
Answer:
Explanation:
Given
acceleration of system a =1.2 m/s^2
Normal Force N=4.45 N
Force exerted F=20 N
Thus
-------1
Normal reaction 
therefore 
Ball will hit the ground after a time of 1.296 s
Explanation:
initial velocity of ball= Vi=0
g= 9.8 m/s²
height =h= 27 ft=8.23 m
using the kinematic equation
h= Vi t + 1/2 gt²
8.23=0(t) + 1/2 (9.8)t²
t²=1.6796
t=1.296 s
