Weak winds that blow for short periods of time with a short fetch.
Answer:
Explanation:
q = 2e = 3.2 x 10^-19 C
mass, m = 6.68 x 10^-27 kg
Kinetic energy, K = 22 MeV
Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A
(a) time, t = 2.8 s
Let N be the alpha particles strike the surface.
N x 2e = q
N x 3.2 x 10^-19 = i t
N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8
N = 2.36 x 10^12
(b) Length, L = 16 cm = 0.16 m
Let N be the alpha particles
K = 0.5 x mv²
22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²
v² = 1.054 x 10^15
v = 3.25 x 10^7 m/s
So, N x 2e = i x t
N x 2e = i x L / v
N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)
N = 4153.85
(c) Us ethe conservation of energy
Kinetic energy = Potential energy
K = q x V
22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V
V = 1.17 x 10^7 V
To calculate the velocity of the sound wave, we use this formula:
V = 331 + [0.6*T],
Where V is the velocity and T represents temperature.
When the temperature is 36 degree Celsius, we have
V = 331 + [0.6 * 36]
V = 331 + 21.6 = 352.6
Therefore, V = 352.6 m/s.
Answer: 
Explanation:
In the image attached with this answer are shown the given options from which only one is correct.
The correct expression is:

Because, if we derive velocity
with respect to time
we will have acceleration
, hence:

Where
is the mass with units of kilograms (
) and
with units of meter per square seconds
, having as a result 
The other expressions are incorrect, let’s prove it:
This result has units of
This result has units of
This result has units of
and
is a constant
This result has units of
This result has units of
This result has units of
and
is a constant
This result has units of
and
is a constant
because
is a constant in this derivation respect to
This result has units of
and
is a constant