Answer:
(A) = 3.57 m
Explanation:
from the question we are given the following:
diameter (d) = 3.2 m
mass (m) == 42 kg
angular speed (ω) = 4.27 rad/s
from the conservation of energy
mgh = 0.5 mv^{2} + 0.5Iω^{2} ...equation 1
where
Inertia (I) = 0.5mr^{2}
ω = \frac{v}{r}
equation 1 now becomes
mgh = 0.5 mv^{2} + 0.5(0.5mr^{2})(\frac{v}{r})^{2}
gh = 0.5 v^{2} + 0.5(0.5)(v)^{2}
4gh = 2v^{2} + v^{2}
h = 3v^{2} ÷ 4 g .... equation 2
from ω = \frac{v}{r}
v = ωr = 4.27 x (3.2 ÷ 2)
v = 6.8 m/s
now substituting the value of v into equation 2
h = 3v^{2} ÷ 4 g
h = 3 x (6.8)^{2} ÷ (4 x 9.8)
h = 3.57 m
Work done in moving a proton = potential difference×Charge of a proton
= 164×1.6×10⁻¹⁹ = 2.624×10⁻¹⁷ J
This work done should be equal to change in kinetic energy.
Initial speed of proton is zero therefore K.E initial will be zero.
Work done = final kinetic energy = 2.624×10⁻¹⁷ J
K.E = mv²/2
v² = 2(2.624×10⁻¹⁷)/1.6×10⁻²⁷ = 3.28×10¹⁰ m/s
∴ v = 1.811×10⁵ m/s
Hey There!
The angle of incidence increases and the angle of refraction increases.
Thank You!
Answer:
The period of the resulting oscillatory motion is 0.20 s.
Explanation:
Given that,
Spring constant
We need to calculate the time period
The object is at rest and has no elastic potential but it does has gravitational potential.
If the object falls then the the gravitational potential change in to the elastic potential.
So,
Where,h = distance
k = spring constant
Put the value into the formula
Using formula of time period
Put the value into the formula
Hence, The period of the resulting oscillatory motion is 0.20 s.
Answer:
The amount of charge on the plates increases during this process
Explanation:
In order to maintain a constant voltage across the plates, pushing the plates closer will increase the charge on the plate