Answer:
Second projectile is 1.4 times faster than first projectile.
Explanation:
By linear momentum conservation
Pi = Pf
m x U + M x 0 = (m + M) x V
Now Since this projectile + pendulum system rises to height 'h', So using energy conservation:
KEi + PEi = KEf + PEf
PEi = 0, at reference point
KEf = 0, Speed of system zero at height 'h'
PEf = (m + M) g h
So,
So from above value of V
Initial velocity of projectile =U
Now Since mass of projectile and pendulum are constant, So Initial velocity of projectile is proportional to the square root of height swung by pendulum.
Which means
U₂ = 1.41 U₁
Therefore we can say that ,Second projectile is 1.4 times faster than first projectile.
Answer and Explanation:
In optics, a CoC(Circle of Confusion) is defined the minimum cross section of a paraxial bundle of rays made by a lens which is sphero-cylindrical type and can be viewed as an optical spot, which do not converge perfectly at the focus while a point source is being imaged due to spherical aberration.
The Circle of Confusion is also referred to as circle of indistinctness or a blur spot
Answer:
The voltage is 
Explanation:
From the question we are told that
The time that has passed is 
Here 
is know as the time constant
The voltage of the power source is 
Generally the voltage equation for charging a capacitor is mathematically represented as
![V = V_b [1 - e^{- \frac{t}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7Bt%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Cfrac%7B%5Ctau%7D%7B2%7D%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\tau}{2\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Ctau%7D%7B2%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{1}{2} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B1%7D%7B2%7D%20%7D%5D)
=>
Answer:
Explanation:
Given equation is ,
x =t + 2 t³ ,
dx/dt = velocity ( v ) = 1 + 6 t²
a) kinetic energy = 1/2 m v² = .5 x 4 ( 1 + 6 t² )² = 2 ( 1 + 6 t²)²
b ) Acceleration = dv /dt = 12 t .
force( F ) = mass x acceleration = 4 x 12 t = 48 t
Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )
work done = ∫ F dx =∫ 48 t x( 1 + 6t² )dt ; = [48t²/2 + 48 x 6 x t³ /3 = 24 t² + 96 t³ )]₀² = 864 J
