Find the angles which the vecotr <span><span><span><span><span>v </span><span>⃗ </span></span>=3i−6j+2k</span> </span><span><span>v→</span>=3i−6j+2k</span></span>
makes with the coordinate axes.
If the angles are <span><span><span>α,β,θ</span> </span><span>α,β,θ</span></span>
, show that for any 3-dimensional vector:
. . . <span><span><span><span><span>cos </span><span>2 </span></span>α+co<span><span>s </span><span>2 </span></span>β+<span><span>cos </span><span>2 </span></span>θ=1</span> </span><span><span>cos2</span>α+co<span>s2</span>β+<span>cos2</span>θ=1</span></span>
The Answer Is Number Of Neutrons!
Answer:
kinetic energy (K.E) = 5.28 ×10⁻¹⁷
Explanation:
Given:
Mass of α particle (m) = 6.50 × 10⁻²⁷ kg
Charge of α particle (q) = 3.20 × 10⁻¹⁹ C
Potential difference ΔV = 165 V
Find:
kinetic energy (K.E)
Computation:
kinetic energy (K.E) = (ΔV)(q)
kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)
kinetic energy (K.E) = 528 (10⁻¹⁹)
kinetic energy (K.E) = 5.28 ×10⁻¹⁷
Answer:
A time that is less than half an hour
Explanation:
it says velocity is positive so it would yield to a negative acceleration
