Answer:
A u = 0.36c B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
u’ (1- u v / c²) = u -v
u + u ’uv / c² = v - u’
u (1 + u ’v / c²) = v - u’
u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
u = 0.51c / (1 + 0.4042)
u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
u = 1.29c / (1- 0.329)
u = 0.961c
The Specific Heat Capacity of silver is 230 J/kgK, melting point is 961.8 C so the difference is 941.8K. Now we simply do q=230J/kgK*16.5kg*941.8K and that is 3 574 131 J
Given data
*The given mass of the pendulum is m = 3 kg
*The given height is h = 0.3 m
The formula for the maximum speed of the pendulum is given as
![v_{\max }=\sqrt[]{2gh}](https://tex.z-dn.net/?f=v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B2gh%7D)
*Here g is the acceleration due to the gravity
Substitute the values in the above expression as
![\begin{gathered} v_{\max }=\sqrt[]{2\times9.8\times0.3} \\ =2.42\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B2%5Ctimes9.8%5Ctimes0.3%7D%20%5C%5C%20%3D2.42%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
Hence, the maximum speed of the pendulum is 2.42 m/s
Answer:
Explanation:
given,
side of square loop = a = 2.10 cm
Resistance of the wire = 1.30×10⁻² Ω
Length of the loop = c = 1.10 cm
rate of increasing current = 130 A/s
At the top of the height, the velocity is zero and acceleration is negative of acceleration due to gravity ( i.e 
).
The time of the ball in air is 3.2 s, so ascending time is 
.
Therefore from kinematic equation,
Substituting the values we get,
, Here v = 0 at top.
Now from equation,
, here h is the height .
So,
.
Thus, the ball reached at its maximum height of 12.48 m.
