3q+1 > 1
solve please
Answers
Step-by-step explanation:
Let the positive integer n is of the form 3q,3q+1, and 3q+2
If n=3q
Squaring both sides, we get,
=>n
2
=9q
2
=>n
2
=3(3q
2
)
=>n
2
=3m, where m=3q
2
Now, if n=3q+1
=>n
2
=(3q+1)
2
=>n
2
=9q
2
+6q+1
=>n
2
=3q(3q+2)+1
=>n
2
=3m+1, where m=q(3q+2)
Now, if n=3q+2
=>n
2
=(3q+2)
2
=>n
2
=9q
2
+12q+4
=>n
2
=3q(3q+4)+4
=>n
2
=3q(3q+4)+3+1
=>n
2
=3m+1 where m=(3q
2
+4q+1)
Hence, n
2
integer is of the form 3m and 3m+1 not 3m+2
Answer:
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C