Answer:
451.13 J/kg.°C
Explanation:
Applying,
Q = cm(t₂-t₁)............... Equation 1
Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.
Make c the subject of the equation
c = Q/m(t₂-t₁).............. Equation 2
From the question,
Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C
Substitute these values into equation 2
c = 1500/[0.133(45-20)]
c = 1500/(0.133×25)
c = 1500/3.325
c = 451.13 J/kg.°C
Answer:
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Answer:
3540.5N
Explanation:
Step one:
given data
mass m= 0.196kg
speed v= 31m/s
distance r= 5.32cm = 0.0532m
Step two
The expression relating force, mass, velocity and distance is
F= mv^2/r
substitute we have
F=0.196*31^2/0.0532
F=0.196*961/0.0532
F=188.356/0.0532
F=3540.5N
Answer:
You would have to find the friction force of the rubber block which would be found with the equation of Normal force (mass*gravity) times cooeficient of friction which would give 8.82 N for the amount of friction and because you need more force than 8.82 N (assuming gravity is 9.8)
Answer:
4.1666666 seconds
Explanation:
100 divided by 24 will give you about 4.1666666 seconds or 4 1/6 seconds. Hope it helps!
