Answer:
3.16X10∧-11 m
Explanation:
1/2 mv2 = qV (KE = Electric potential energy)
velocity = √2qV/m = √( 2X 1.6X10∧-19 X 1500/9.11X10∧-31)
2.3X10∧7m/s
now use De Broglie equation
λ = h/mv
= 6.62X10∧-34/( 9.11X10∧-31 X 2.3X10∧7)
3.16 X 10∧-11 m
or
use the above equations and substitute to get the final eqiation
λ = h/√(2mqV) = 3.16X 10∧-11 m
Answer:
I think it's option D
Explanation:
I think it's option D but not so sure
Explanation:
C. neutron.
it does not contain ionizing characteristics.
hope it helps. :)
Answer:
(A) 0.2306 m
(B) 1.467 Hz
(C) 0.1152 m
Explanation:
spring constant (K) = 16.4 N/m
mass (m) = 0.193 kg
acceleration due to gravity (g) = 9.8 m/s^{2}
(A) force = Kx, where x = extension
mg = Kx
0.193 x 9.8 = 16.4x
x = 0.1153 m
now the mass actually falls two times this value before it gets to its equilibrium position ( turning point ) and oscillates about this point
therefore
2x = 0.2306 m
(B) frequency (f) = \frac{1}{2π} x 
frequency (f) = \frac{1}{2π} x 
frequency = 1.467 Hz
(C) the amplitude is the maximum position of the mass from the equilibrium position, which is half the distance the mass falls below the initial length of the spring
= \frac{0.2306}{2} = 0.1152 m
Answer:
0.8 m
Explanation:
Draw a free body diagram. There are three forces:
Weight force mg pulling down,
Normal force N pushing up,
and friction force Nμ pushing towards the center.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
Sum of forces in the centripetal direction:
∑F = ma
Nμ = m v²/r
Substitute and simplify:
mgμ = m v²/r
gμ = v²/r
Write v in terms of ω and solve for r:
gμ = ω²r
r = gμ/ω²
Plug in values:
r = (10 m/s²) (0.5) / (2.5 rad/s)²
r = 0.8 m
