Answer:
Minimum number of photons required is 1.35 x 10⁵
Explanation:
Given:
Wavelength of the light, λ = 850 nm = 850 x 10⁻⁹ m
Energy of one photon is given by the relation :
....(1)
Here h is Planck's constant and c is speed of light.
Let N be the minimum number of photons needed for triggering receptor.
Minimum energy required for triggering receptor, E₁ = 3.15 x 10⁻¹⁴ J
According to the problem, energy of N number of photons is equal to the energy required for triggering, that is,
E₁ = N x E
Put equation (1) in the above equation.
Substitute 3.15 x 10⁻¹⁴ J for E₁, 850 x 10⁻⁹ m for λ, 6.6 x 10⁻³⁴ J s for h and 3 x 10⁸ m/s for c in the above equation.
N = 1.35 x 10⁵
Answer:
a
b
Explanation:
From the question we are told that
The mass of the rock is 
The length of the small object from the rock is 
The length of the small object from the branch 
An image representing this lever set-up is shown on the first uploaded image
Here the small object acts as a fulcrum
The force exerted by the weight of the rock is mathematically evaluated as
substituting values
So at equilibrium the sum of the moment about the fulcrum is mathematically represented as
Here 
is very small so 
and 
Hence
=> 
substituting values
The mechanical advantage is mathematically evaluated as
substituting values
The correct answer is letter b.
To find the answer follow the following steps.
1. 6524.96 x .25 = X
2. 1631.24 = X
This works for all of the given answers to find the correct answer.
Lysosomes are used by the cell to digest or breakdown multifaceted organic molecules
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so 
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
