Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
![W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7BL%7D%20Fdx%20%3D%20%5Cint_%7B0%7D%5E%7BL%7Dax%20%5C%2C%20dx%20%3D%20%5Cfrac%7Ba%7D%7B2%7D%20%20%5Bx%5E%7B2%7D%20%5D_%7B0%7D%5E%7BL%7D%20%3D%20%20%5Cfrac%7BaL%5E%7B2%7D%7D%7B2%7D%20)
Answer:
As it is given that Bulk modulus and density related to velocity of sound
by rearranging the equation we can say
now we need to find the SI unit of Bulk modulus here
we can find it by plug in the units of density and speed here
so SI unit will be
SO above is the SI unit of bulk Modulus
Answer:
option C
Explanation:
given,
Force by the engine on plane in West direction = 350 N
Frictional force on the runway = 100 N in east
force exerted by the wind = 100 N in east
net force and direction = ?
consider west to be positive and east be negative.
when airplane will be moving there will be frictional as well as wind resistance will be acting in opposite direction of airplane
Net force = 350 N - 100 N - 100 N
= 150 N
as our answer comes out to be positive so the airplane will be moving in West
hence, the correct answer is option C
Answer: To focus on a near object – the lens becomes thicker, this allows the light rays to refract (bend) more strongly. To focus on a distant object – the lens is pulled thin, this allows the light rays to refract slightly.
Explanation:
B) They are all sources of electricity
Hope it helps!!
