Potential energy is a relative measure, so the answer is dependent on the assumptions we make. The potential energy in the car is going to be gravitational potential energy(PE). PE = mgh, where m is the mass, g is 9.8 m/s^2, and h is the height. So PE = 2000*9.8*h = 19600h. The final answer obviously depends on h. Most likely the problem is assuming that 30 meters under the top of the hill is considered 0 meters. Then h would be 30m and PE would equal 588 kJ.
In projectile motion we can say initial components of velocity is
now here horizontal motion is uniform motion as there is no force in this direction while in vertical direction due to gravity it will accelerate
so we have
so here we have x and y coordinates are related to each other as it will take same time to move the positions
so here correct options are
<em>Projectile motion is a combination of horizontal and vertical motion.
</em>
<em>The horizontal and vertical motions of a projectile are dependent on each other.</em>
Collagenous fibers - Purpose : Bind bones and other tissues to each other<span>
Elastic Fibers : - Purpose : </span>Allow organs like arteries and lungs to recoil<span>
Reticular Fiber : - Purpose : </span><span>Form a scaffolding for other cells
Hope this helps ! <3 :)</span>
Answer:
Linear mass density,
Explanation:
Given that,
Mass of the string, m = 0.3 g = 0.0003 kg
Length of the string, l = 1.5 m
The linear mass density of a string is defined as the mass of the string per unit length. Mathematically, it is given by :
or
So, the linear mass density of a string is 
. Hence, this is the required solution.
Answer:
The thermal conductivity of the wall = 40W/m.C
h = 10 W/m^2.C
Explanation:
The heat conduction equation is given by:
d^2T/ dx^2 + egen/ K = 0
The thermal conductivity of the wall can be calculated using:
K = egen/ 2a = 800/2×10
K = 800/20 = 40W/m.C
Applying energy balance at the wall surface
"qL = "qconv
-K = (dT/dx)L = h (TL - Tinfinity)
The convention heat transfer coefficient will be:
h = -k × (-2aL)/ (TL - Tinfinty)
h = ( 2× 40 × 10 × 0.05) / (30-26)
h = 40/4 = 10W/m^2.C
From the given temperature distribution
t(x) = 10 (L^2-X^2) + 30 = 30°
T(L) = ( L^2- L^2) + 30 = 30°
dT/ dx = -2aL
d^2T/ dx^2 = - 2a
