Nearly equal the output work is greater than the input work because of friction.All machines use some amount of input work to overcome friction.The only way to increase the work output is to increase the work you put into the machine.You cannot get more work out of a machine than you put into it.
A. the light bulb goes out once the circuit is open since it causes the flow of electricity to cut off. the light bulb dosent get the energy it needs to light up
Explanation:
B. a simple example of this in our every day life is a light switch. when you switch the light on then the circuit is closed and the energy transfers to the light bulb, when u switch the light off then you cut off the lights source of energy which causes the light to turn off.
Answer:
0.12 K
Explanation:
height, h = 51 m
let the mass of water is m.
Specific heat of water, c = 4190 J/kg K
According to the transformation of energy
Potential energy of water = thermal energy of water
m x g x h = m x c x ΔT
Where, ΔT is the rise in temperature
g x h = c x ΔT
9.8 x 51 = 4190 x ΔT
ΔT = 0.12 K
Thus, the rise in temperature is 0.12 K.
Does it have to be that exact word. cause it is just another term for psuedopodium
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
