Answer:
a)the arrow will have a very little kinetic energy after the archer and its realesead the string in the picture of the right
Explanation: you are at my same school and I pass it correct also as shown in the image it should be like that
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
Calvin Cycle is the part of the process of photosynthesis, by which Plant makes Food (Glucose), it is independent of the light. It is essential as it is the primary source of energy on the earth
Hope this helps!
Answer:
axial V = 0
equatorial V = k q 2a / (x² -a²), V = k q 2x / (a² -x²)
Explanation:
A dipole is a system formed by two charges of equal magnitude, but different sign, separated by a distance 2a; let's look for the electrical potential in an axial line
V = k (q / √(a² + y²) - q /√ (a² + y²))
V = 0
the potential on the equator
we place the positive charge to the left and perform the calculation for a point outside the dipole
V = k (q / (x-a) - q / (x + a))
V = k q 2a / (x² -a²)
we perform the calculation for a point between the dipo charges
V = k (q / (a-x) - q / (a + x))
V = k q 2x / (a² -x²)
