The amount of air resistance<span> an </span>object<span> experiences depends on its speed, its cross-sectional area, its shape and the density of the </span>air<span>. </span>Air<span> densities vary with altitude, temperature and humidity. Nonetheless, 1.29 kg/m</span>3<span> is a very reasonable value. The shape of an </span>object affects<span> the drag coefficient (C</span>d<span>)</span>
Answer:
Explanation:
Given
Mass of object (m)=6 kg
falling height(h)=10 m
mass of water(
)=600 gm
temperature of water =15
specific heat of water 
Let T be the Final Temperature of water
Here Object Potential Energy is converted into Heat energy which will be absorbed by water
Potential Energy(P.E.)
Heat supplied
T-16=0.234
This is not an efficient way of heating water as there is only
increase in temperature.
Force [kgms^-2] = mass [kg] x acceleration [ms^-2]
Work = force x distance
Work = [kgms^-2] x [m]
Work = [kgm^2s^-2]
Answer:
Tt = 70 + 135e^-0.031t
13 minutes
Explanation:
Given that :
Initial temperature, Ti = 205°
Temperature after 2.5 minutes = 195°
Temperature of room, Ts= 70
Using the relation :
Tt = Ts + Ce^-kt
Temperature after time, t
When freshly poured, t = 0
205 = 70 + Ce^-0k
205 = 70 + C
C = 205 - 70 = 135°
T after 2.5 minutes to find proportionality constant, k
Tt = Ts + Ce^-kt
195 = 70 + 135e^-2.5k
125 = 135e^-2.5k
125 / 135 = e^-2.5k
0.9259 = e^-2.5k
Take In of both sides :
−0.076989 = - 2.5k
k = −0.076989 / - 2.5
k = 0.031
Equation becomes :
Tt = 70 + 135e^-0.031t
t when Tt = 160
160 = 70 + 135e^-0.031k
90 = 135e^-0.031t
90/135 = e^-0.031t
0.6667 = e^-0.031t
In(0.6667) = - 0.031t
−0.405465 = - 0.031t
t = 0.405465/ 0.031
t = 13.071
t = 13 minutes
