Volume=mass/density
volume=455.6/19.3
volume=23.6 mL
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
Answer:
The speed stays constant after the force stops pushing.
Explanation:
Speed always stays constant when the force stops pushing it.
Answer:
16.33°C
Explanation:
Applying,
Heat lost by copper = heat gained by water
cm(t₁-t₃) = c'm'(t₃-t₂).............. Equation 1
Where c = specific heat capacity of copper, m = mass of copper, c' = specific heat capacity of water, m' = mass of water, t₁ = initial temperature of copper, t₂ = initial temperature of water, t₃ = final equilibrium temperature.
From the question,
Given: m = 50 kg, t₁ = 140°C, m' = 90 L = 90 kg, t₂ = 10°C
Constant: c = 385 J/kg°C, c' = 4200J/kg°C
Substitute these values into equation 1
50(385)(140-t₃) = 90(4200)(t₃-10)
(140-t₃) = 378000(t₃-10)/19250
(140-t₃) = 19.64(t₃-10)
140-t₃ = 19.64t₃-196.6
19.64t₃+t₃ = 196.4+140
20.64t₃ = 336,4
t₃ = 336.4/20.6
t₃ = 16.33°C
