= (18 x 10^-6 /°C)(0.125 m)(100° C - 200 °C)
= -0.00225 m
New length = L + ΔL
= 1.25 m + (-0.00225 m)
= 1.248
D
Explanation:
Let magnitude of the two forces be x and y.
Resultant at right angle R1= √15N) and at
60 degrees be R2= √18N.
Now, R1 = √(x² + y²) = √15,
R2= √(x² + y² +2xycos50) = √18.
So x² + y² = 15,
and x² + y² + 1.29xy = 18,
therefore 1.29xy = 3,
y = 3/1.29x.
y = 2.33/x
Now, x2 + (2.33/x)2 = 15,
x² + 5.45/x² = 15
multiply through by x²
x⁴ + 5.45 = 15x²
x⁴ - 15x2 + 5.45 = 0
Now find the roots of the equation, and later y. The two values of x will correspond to the
magnitudes of the two vectors.
Good luck
Answer:
the mass of water is 0.3 Kg
Explanation:
since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:
Q water + Q copper = Q surroundings =0 (insulated)
Q water = - Q copper
since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature
and denoting w as water and co as copper :
m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) = m co * c co * (T co - Ti eq)
m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]
We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C
if we assume that both specific heats do not change during the process (or the change is insignificant)
m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]
m w= 1.80 kg * 0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))
m w= 0.3 kg
