Answer:
speed of electrons = 3.25 × 
m/s
acceleration in term g is 3.9 × 
g.
radius of circular orbit is 2.76 × 
m
Explanation:
given data
voltage = 3 kV
magnetic field = 0.66 T
solution
law of conservation of energy
PE = KE
qV = 0.5 × m × v²
v = 
v = 
v = 3.25 × m/s
and
magnetic force on particle movie in magnetic field
F = Bqv
ma = Bqv
a = 
a = 
a = 3.82 × 
m/s²
and acceleration in term g
a = 
a = 3.9 × g
acceleration in term g is 3.9 × g.
and
electron moving in circular orbit has centripetal force
F = 
Bqv =
r = 
r = 
r = 2.76 × m
radius of circular orbit is 2.76 × m
Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
The correct answer should be bounced off since the light ray hit the surface and reflected towards a new location. It therefore bounced off.
Answer:
Δx = 1.2 m
Explanation:
The CHANGE of spring length) (Δx) can be found using PS = ½kΔx²
Δx = √(2PS/k) = √(2(450)/650) = 1.17669... ≈ 1.2 m
The actual length of the spring is unknown as it varies with material type, construction method, extension or compression, and other variables we have no clue about.
