Answer:
Angle with the +x axis is θ = 79.599degree
Then the velocity of owner = 1.235m/s
Explanation:
Given that the mass of dog is m1 =26.2 kg
velocity of dog is u1 = 3.02 m/s (north)
mass of cat is m2 = 5.3 kg
velocity is u2 = 2.74 m/s (east )
Mass of owner is M = 65.1 kg
Consider the east direction along +x axis andnorth along +y
momentum of dog is Py = m1 x u1
= 79.124 kg.m/s (j)
momentum of cat is Px = m2 x u2
= 14.522 kg.m/s (i)
Then the net magnitude of momentum is P = (Px2 + Py2)1/2
= 80.445
Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree
Then the velocity of owner is v = P / M = 1.235 m/s
MEMORIZED E=h*v h=6.626x10-34J*s INFORMED v=7.21x1014S-1CALCULATE E=h*v E=(6.626x10-34J*s)*(7.21x1014s-1) The "s" cancels out. s-1=1/s so you get s/s so you are left with Solution 4.78 10-19 J OR .478 aJ <span>Apex - 467 nm ^.^ hopefully thats the correct thing</span>
Answer:
A.
Explanation:
NEAR THE CENTER OF TECTONIC PLATES.
Answer:
Vb = k Q / r r <R
Vb = k q / R³ (R² - r²) r >R
Explanation:
The electic potential is defined by
ΔV = - ∫ E .ds
We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product
VB - VA = - ∫ E dr
Let's substitute every equation they give us and we find out
r> R
Va = - ∫ (k Q / r²) dr
-Va = - k Q (- 1 / r)
We evaluate with it Va = 0 for r = infinity
Vb = k Q / r r <R
We perform the calculation of the power with the expression of the electric field that they give us
Vb = - int (kQ / R3 r) dr
We integrate and evaluate from the starting point r = R to the final point r <R
Vb = ∫kq / R³ r dr
Vb = k q / R³ (R² - r²)
This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity
Complete Question
The complete question is shown on the first uploaded image
Answer:
The theoretical angular magnification lies within the angular magnification range
Explanation:
From the question we are told that
The focal length of B is 
The focal length of A is 
The theoretical angular magnification is mathematically represented as


Form the question the measured angular magnification ranges from 4 -5
So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range