Answer:
c = 1163.34 J/kg.°C
Explanation:
Specific heat capacity:
"Specific heat capacity is the amount of heat energy required to raise the temperature of a substance per unit of mass. The specific heat capacity of a material is a physical property."
Use this equation:
mcΔT = ( mw c + mAl cAl ) ΔT'
Rearranging the equation to find the specific heat (c) you get this:
c = (( mw c + mAl cAl ) ΔT') / (mΔT)
c = (( 0.285 (4186) + (0.15)(900)) (32 -25.1)) / ((0.125) (95 - 32))
c = 1163.34 J/kg.°C
Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf, 
Magnetic Field, B = 0.10 T
Hall emf, 
Now,
Drift velocity, 
Now, the expression for the electric field is given by:
(1)
And
Thus eqn (1) becomes
where
d = distance
(2)
(a) When 
(b) When 
Answer:
1.06 secs
Explanation:
Initial speed of sled, u = 8.4 m/s
Final speed of sled, v = 5.8 m/s
Coefficient of kinetic friction, μ = 0.25
Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:
FΔt = mv - mu
where m = mass of the object
Δt = time interval
F = force applied
The force applied on the sled is the frictional force, which is given as:
F = -μmg
where g = acceleration due to gravity
Therefore:
-μmgΔt = mv - mu
-μmgΔt = m(v - u)
-μgΔt = v - u
Making Δt subject of formula:
Δt = (v - u) / -μg
Δt = (5.8 - 8.4) / (-0.25 * 9.8)
Δt = -2.6/ -2.45
Δt = 1.06 secs
It took the sled 1.06 secs to travel from A to B.
A fuse melts to protect a circuit.
