a wildlife biologist examines for a genetic trait he suspects may be linked to sensitivity to industrial toxins in the environme
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The period of a simple pendulum is given by:
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
(1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
And its period is the reciprocal of its frequency:
So now we can use eq.(1) to find the gravitational acceleration of the planet:
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
And its period is the reciprocal of its frequency:
So now we can use eq.(1) to find the gravitational acceleration of the planet:
Answer:
λ = 5.65m
Explanation:
The Path Difference Condition is given as:
δ= ;
where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.
m = no of openings which is 2
∴δ=
n is the index of refraction of the medium in which the wave is traveling
To find δ we have;
δ=
δ=
δ=
δ=
δ=
δ=
δ= 82.15 -73.68
δ= 8.47
Again remember; to calculate the wavelength of the ocean waves; we have:
δ=
δ= 8.47
8.47 =
λ =
λ = 5.65m
Answer:
v=77.62 m/s
Explanation:
Given that
h= - 300 m
speed of the bird ,u= 5 m/s
Lets take Speed of the berry when it hit the ground = v m/s
we know that ,if object is moving upward
v² = u² - 2 g h
u=Initial speed
v=Final speed
h=Height
Now by putting the values
v² = u² - 2 g h
v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)
v² =25 + 20 x 300
v² ==25 + 6000
v² =6025
v=77.62 m/s
Therefore the final speed of the berry will be 77.62 m/s.
