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:
Explanation:
Component of force perpendicular to stick
= F Sin 60°
=√3 / 2 F.
Taking torque about the other end
= √3 / 2 F x 1 Nm
Weight of stick = 60 gm
= 60 x 10⁻³ kg
= 60 x 10⁻³ x 9.8 N
= .588 N
This weight will act from the middle point of stick so torque about the
other end
= .588 x 1 Nm
Balancing these two torques we have
.588 = √3 /2 F
F = 0.679 N
Answer:
In both cases, energy will move from an area of higher temperature to an area of lower temperature. So, the energy from room-temperature air will move into the cold water, which warms the water.
Explanation:
kinematic equation
v=u+at
v-u=at
v-u = 1x5
the driver will have increased speed by 5 m/s. actual speeds unknown
