Answer:
(1) tropical storm
(2) Severe tropical storm
(3) 305 kmh
Answer:
11.5 m
Explanation:
First, find the time it takes to land.
Given:
Δy = -1.5 m
v₀ = 10 sin 37° m/s
a = -9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
-1.5 m = (10 sin 37° m/s) t + ½ (-9.8 m/s²) t²
-1.5 = 6.02 t − 4.9 t²
4.9 t² − 6.02 t − 1.5 = 0
Solve with quadratic formula:
t = [ -(-6.02) ± √((-6.02)² − 4(4.9)(-1.5)) ] / 2(4.9)
t = 1.44
Now find how far it moves horizontally in that time:
Given:
t = 1.44 s
v₀ = 10 cos 37° m/s
a = 0 m/s²
Find: Δx
Δx = v₀ t + ½ at²
Δx = (10 cos 37° m/s) (1.44 s) + ½ (0 m/s²) (1.44 s)²
Δx = 11.5 m
The cat lands 11.5 meters from the table.
Answer:
(1)
(2)
but
transition not allowed.
Explanation:
Atoms can be described by the quantum number n, spin quantum number S, angular momentum quantum number L, and total angular momentum quantum number J. Based on approximation Russel- Saunders electron coupling, the atomic term symbol can be written as
.
The conditions or selection rule to promoting the electron are discussed below:
(1) The total spin should not change that is
.
(2) The total angular momentum change should be,
but
transition not allowed.
Answer:
2.5 kg/m³
Explanation:
Absolute pressure = gauge pressure + atmospheric pressure
P = Pg + Pa
The gauge pressure caused by the weight of a fluid is called the static pressure. It is equal to the density of the fluid × acceleration due to gravity × depth of the fluid,
Pg = ρgh
Therefore:
P = ρgh + Pa
300 Pa = ρ (10 m/s²) (8.00 m) + 100 Pa
ρ = 2.5 kg/m³
Explanation:
It is given that,
A helicopter blade spins at exactly 100 revolutions per minute.
Its tip is 5.00 m from the center of rotation, r = 5 m
(a) Let v is the average speed of the blade tip in the helicopter’s frame of reference. Distance covered by the helicopter, 
In 100 revolutions, 
So, average speed of the blade tip in one second is given by :


v = 52.35 m/s
(b) The average velocity over one revolution is zero because the net displacement in one rotation is 0.
Hence, this is the required solution.