Answer:
The velocity of the ball after 5 seconds will be 49 m/s
Explanation:
<em>v = final velocity</em>
<em>u = initial velocity</em>
<em>g = acceleration due to gravity</em>
<em>t = time</em>
Initial velocity of the ball = 0 (As the ball is dropped from rest )
Acceleration due to gravity = 9.8 m/s
Time taken = 5 sec
As the acceleration due to gravity is constant in both the cases we can use the equations of motion in order to solve this question
Part I :- As we already know the values of u,g,ant t we can use the first equation of motion in order to find v
Part II :- As we know the values of u, t , g we can use the second equation of motion in order to find s.
Velocity of the ball after 5 seconds
Distance covered by the ball in 5 sec
Answer:
The answer is below
Explanation:
The question is not complete since the liquid density is not given.
Archimedes principle states that a body at rest in a fluid is acted upon by an upward force known as the buoyant force. The buoyant force is equal to the weight of the fluid displaced.
An object floats when it is placed in a liquid only if the density of the object is less than the density of the liquid. Therefore those metals with density less than that of mercury would float while those with density greater than mercury would sink.
Answer:
λ = 396.7 nm
Explanation:
For this exercise we use the diffraction ratio of a grating
d sin θ = m λ
in general the networks works in the first order m = 1
we can use trigonometry, remembering that in diffraction experiments the angles are small
tan θ = y / L
tan θ = 
= sin θ
sin θ = y / L
we substitute
= m λ
with the initial data we look for the distance between the lines
d = 
d = 1 656 10⁻⁹ 1.00 / 0.600
d = 1.09 10⁻⁶ m
for the unknown lamp we look for the wavelength
λ = d y / L m
λ = 1.09 10⁻⁶ 0.364 / 1.00 1
λ = 3.9676 10⁻⁷ m
λ = 3.967 10⁻⁷ m
we reduce nm
λ = 396.7 nm
Answer:
I am not really sure but i think its option 2
Explanation:
Answer:
the answer to your question is 4 cm long
Explanation:
