Spinning a marshmallow over a fire is effective maybe if you hang it over the fire and heat it up equally on each side
In theory, yes. The 2 problems are the materials used for clinical thermometers, & the temperature capacity of the clinical thermometer. If anything, change the material & extend the measurement threshold. At that point, it wouldn´t be used for clinical garbage anymore.
<span>The answer is The conductance of a conductor is inversely
proportional to the cross-sectional area of the conductor.</span>
<span>Conductance is directly related to the ease offered by any material to the passage of electric current. Conductance is the opposite of resistance. The higher the conductance, the lower the resistance and vice versa, the greater the resistance, the less conductance, so both are inversely proportional</span>
Answer:
is the time taken by the car to accelerate the desired range of the speed from zero at full power.
Explanation:
Given:
Range of speed during which constant power is supplied to the wheels by the car is 
.
- Initial velocity of the car,
- final velocity of the car during the test,
- Time taken to accelerate form zero to 32 mph at full power,
- initial velocity of the car,
- final desired velocity of the car,
Now the acceleration of the car:
Now using the equation of motion:
is the time taken by the car to accelerate the desired range of the speed from zero at full power.
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
