Answer:
The wavelengths of C1 is 10.4m, A6 is 0.193m and B7 is 0.0861m
Explanation:
Using the formula V = f×λ . Then substitute the following values into the formula:
a) v=340m/s
f=32.7 Hz
λ=V ÷ f
= 340 ÷ 32.7
= 10.4m (3s.f)
b) λ=340 ÷ 1760
= 0.193m (3s.f)
c) λ=340÷3951.1
= 0.0861m (3s.f)
(Correct me if I am wrong)
Answer:
m/s
Explanation:
Assumptions: 0° is true North, and 90° is east (along the x-axis).
To solve this problem we must use the expression:
Where 
is the velocity in the y-direction (East), 
is the total velocity in the direction which the aircraft is travelling, and 
is the direction the aircraft is travelling (angle from the y-axis).
Using the equation above, we obtain the y-component of velocity
m/s which is rounded to 227 m/s (due to the number of significant figures in the question).
The Balmer light series comes under the visible light.
<u>Explanation:</u>
The transition of electrons from higher to energy level with 2 as principal quantum number results in the spectral emission lines of hydrogen atom and this series of lines are known as Balmer series.
Mostly, these lines has the wavelength of more than 400 nm but lesser than 700 nm. Generally of the four categories namely, 410, 434, 486, 656 nm which comes under the type of visible light. So, it can be concluded that the Balmer series light falls under visible light.
In astronomy, Balmer lines occur in various stellar (celestial or astronomical) objects due to the higher content of hydrogen in the universe. Therefore, they are commonly seen and relatively strong when compared to other element lines.
Note: nm is nanometer (one billionth of a meter in length)
Answer:
A. 66.0 m/s downwards
Explanation:
The Tower has a height of 444m
The book is dropped ,finding the velocity of the book 222m above the ground, means the book will be on air for a height of 222 m .
Apply the formula for free fall in a horizontal projection as;
h= u²×sin²∅ /2g where
h= maximum height =222m
g= acceleration due to gravity =9.81 m/s²
∅ = projectile angle = 0
u = velocity of the book
Applying the formula as ;
h= u²×sin²∅ /2g
222 = u²/2*9.81
222*19.62 = u²
4355.64 = u²
√4355.64 = u
65.99 m/s = u
66 m/s downwards
F would 0.825 because the coefficient of kinetic friction between the block and the plane surface uk of 0.35
