Density =mass/volume
density= (0.044kg/(11 x10^-6)
4000kg/m^3
Answer:
wavelength
= 437.27 nm
Explanation:
given data
first bright fringe = 2.96 mm
slit separation = 0.325 mm
distance D = 2.20 m
solution
we know that this is double slit experiment
so we apply here Fringe width formula that is
β =
....................1
is Wavelength of light and D is Distance between screen and slit and d is slit width
so put here value and we get
=
= 437.27 ×
m
wavelength
= 437.27 nm
Answer:
The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.
Explanation:
Let u is the initial speed of the launch. Using first equation of motion as :

a=-g

The velocity of the shell at launch and 5.4 s after the launch is given by :

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.
Answer:
4
Explanation:
We are given that

K.E at x=0 m=20 J
K.E at x=3 m=11 J
We have to find the value of c.
By work energy theorem
Work done=Change in kinetic energy
W=
![W=[\frac{cx^2}{2}-x^3]^{3}_{0}](https://tex.z-dn.net/?f=W%3D%5B%5Cfrac%7Bcx%5E2%7D%7B2%7D-x%5E3%5D%5E%7B3%7D_%7B0%7D)






Answer:
Explanation:
The mass of the block is 0.5kg
m = 0.5kg.
The spring constant is 50N/m
k =50N/m.
When the spring is stretch to 0.3m
e=0.3m
The spring oscillates from -0.3 to 0.3m
Therefore, amplitude is A=0.3m
Magnitude of acceleration and the direction of the force
The angular frequency (ω) is given as
ω = √(k/m)
ω = √(50/0.5)
ω = √100
ω = 10rad/s
The acceleration of a SHM is given as
a = -ω²A
a = -10²×0.3
a = -30m/s²
Since we need the magnitude of the acceleration,
Then, a = 30m/s²
To know the direction of net force let apply newtons second law
ΣFnet = ma
Fnet = 0.5 × -30
Fnet = -15N
Fnet = -15•i N
The net force is directed to the negative direction of the x -axis