The whole question is talking about the amplitude of a wave
that's transverse and wiggling vertically.
Equilibrium to the crest . . . that's the amplitude.
Crest to trough . . . that's double the amplitude.
Trough to trough . . . How did that get in here ? Yes, that's
the wavelength, but it has nothing to do
with vertical displacement.
Frequency . . . that's how many complete waves pass a mark
on the ground every second. Doesn't belong here.
Notice that this has to be a transverse wave. If it's a longitudinal wave,
like sound or a slinky, then it may not have any displacement at all
across the direction it's moving.
It also has to be a vertically 'polarized' wave. If it's wiggling across
the direction it's traveling BUT it's wiggling side-to-side, then it has
no vertical displacement. It still has an amplitude, but the amplitude
is all horizontal.
If the velocity is constant then the acceleration of the object is zero.
Thus when we apply the equation
It remains
or equivalent
Answer:
The reagents correspond to substances that react with others, generating a chemical reaction.
Explanation:
Example:
H2 + 02 -> H202
Hydrogen reacts with oxygen in a synthesis reaction, forming a new compound (hydrogen peroxide) called a product.
Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm
Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as
y = R×(mλ/d)
Where y = distance between nth fringes and the center fringe.
m = order of fringe
λ = wavelength of light = 589nm = 589×10^-9m
R = distance between slits and screen = 1.0m
d = distance between slits = 0.25mm = 0.00025m
For distance between the first dark fringe and the center fringe.
This implies that m = 1
y = 1 × 589×10^-9 × 1/0.00025
y = 589×10^-9/0.00025
y = 2,356,000 × 10^-9
y = 2.356 × 10^-3 m = 2.356mm
For the second dark fringe, this implies that m = 2
y = 1 × 2 × 589×10^-9/0.00025
y = 1178 × 10^-9 /0.00025
y = 4,712,000 × 10^-9
y = 4.712 × 10^-3 m = 4.712mm
Answer:
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
Explanation:
initial veetical speed V₀y=0
Horizontal speed Vx = Vx₀= 3.80m/s
Vertical drop height= 3.90m
Let Vy = vertical speed when it got to the water downward.
g= 9.81m/s² = acceleration due to gravity
From kinematics equation of motion for vertical drop
Vy²= V₀y² +2 gh
Vy²= 0 + ( 2× 9.8 × 3.90)
Vy= √76.518
Vy=8.747457
Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below
V= √Vy² + Vx²
V=√3.80² + 8.747457²
V=9.537m/s
The angle can also be calculated as
θ=tan⁻¹(Vy/Vx)
tan⁻¹( 8.747457/3.80)
=66.52⁰
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
