The type of weather modification is intentional weather modification. Intentional in such a way that there are practical methods applied in order to force/alter the weather condition. For example, frost prevention on crops. Farmers normally utilize large fans to get the warmer air from above to mix with the cold air near the ground surface. By doing so, crops are kept from being ruined by frost.
Answer:
Proton: v=0.689 m/s
Neutron: v=0.688 m/s
Electron: v=1265.078 m/s
Alpha particle: v=0.173 m/s
Explanation:
De Broglie equation allows you to calculate the “wavelength” of an electron or any other particle or object of mass m that moves with velocity v:
λ=
h is the Planck constant: 6.626×10⁻³⁴
We know that the wavelength of the particle is 575 nm (575×10⁻⁹m), so we find the velocity v for each particle:
λ=
v=h÷(mλ)
<u>Proton:</u>
m=1.673×10⁻²⁴ g ·
=1.673×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴
÷(1.673×10⁻²⁷ kg×575×10⁻⁹m)
v=0.689 m/s
<u>Neutron:</u>
m=1.675×10⁻²⁴ g ·
=1.675×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴
÷(1.675×10⁻²⁷ kg×575×10⁻⁹m)
v=0.688 m/s
<u>Electron:</u>
m= 9.109×10⁻²⁸ g ·
=9.109×10⁻³¹ kg
v=h÷(mλ)
v=6.626×10⁻³⁴
÷(9.109×10⁻³¹ kg×575×10⁻⁹m)
v=1265.078 m/s
<u>Alpha particle:</u>
m=6.645×10⁻²⁴ g ·
=6.645×10⁻²⁷ kg
v=h÷(mλ)
v=6.626×10⁻³⁴
÷(6.645×10⁻²⁷ kg×575×10⁻⁹m)
v=0.173 m/s
Answer:
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
Explanation:
Given:
Speed of sound in air = 320 m/s
Speed of sound in water = 1600 m/s
Time taken to reach certain distance in air = 2.5 sec
a.
We have to find the distance traveled by sound in air.
Distance = Product of speed and time.
⇒ 
⇒ 
⇒
meters.
b.
Now we have to find how much time the sound will take to travel in water.
⇒ Time = Ratio of distance and speed
⇒ 
⇒
<em> ...distance = 800 m and speed = 1600 m/s</em>
⇒ 
⇒
seconds.
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
Answer:A
Explanation:Find attached picture file for details
Answer:
A) 12.08 m/s
B) 19.39 m/s
Explanation:
A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:
mg(sinθ) – F_k = ma
Where; F_k is frictional force due to kinetic friction given by the formula;
F_k = (μ_k) × F_n
F_n is normal force given by mgcosθ
Thus;
F_k = μ_k(mg cosθ)
We now have;
mg(sinθ) – μ_k(mg cosθ) = ma
Dividing through by m to get;
g(sinθ) – μ_k(g cosθ) = a
a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)
a = -3.71 m/s²
We are told that distance d = 24.0 m and v_o = 18 m/s
Using newton's 3rd equation of motion, we have;
v = √(v_o² + 2ad)
v = √(18² + (2 × -3.71 × 24))
v = 12.08 m/s
B) Now, μ_k = 0.10
Thus;
a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)
a = 1.08 m/s²
Using newton's 3rd equation of motion, we have;
v = √(v_o + 2ad)
v = √(18² + (2 × 1.08 × 24))
v = 19.39 m/s