We apply the following equation
T = 2π * sqrt (L/g)
Where g is the gravity = 9.8 m/s^2
L is the longitude of the pendulum (Height of the tower)
T is the period. (T = 18s)
We find L.............> (T /2π)^2 = L/g
L = g*(T /2π)^2...........> L = 80.428 meters
Answer:
y = 128.0 km
Explanation:
The minimum separation of two objects is determined by Rayleygh's diffraction criterion, which establishes that two bodies are solved if the first minino of diffraction of one coincides with the central maximum of the second, with this criterion the diffraction equation remains
the diffraction equation for the first minimum is
a sin θ = λ
In the case of circular openings, the equation must be solved in polar coordinates, leaving the expression, we use the approximation that the sine of tea is very small.
θ = 1.22 λ / d
d = 15 cm
to find the distance we can use trigonometry
tan θ = y / L
tan θ = sin θ / cos θ = θ
substituting
y / L = λ / d
y = L λ /d
let's calculate
y = 384 10⁸ 500 10⁻⁹ / 0.15
y = 1.28 10⁵ m
Let's reduce to km
y = 1.28 10⁵ m (1km / 10³ m)
y = 128.0 km
the correct answer is 120 km away
100cm to go in 600 secs = 10mins
2 positive, 1 negative .... net 1 positive per step
Answer:0.45ohms
Explanation:
Let R be there equivalent resistance
1/R=1/r+1/r+1/r
1/R=1/5+1/1+1/1
1/R=1/5+2
1/R=(1+10)/5
1/R=11/5
Cross multiplying we get
11R=5
Divide both sides by 11
11R ➗ 11=5 ➗ 11
R=0.45ohms
To solve this problem we will use the kinematic equations of descriptive motion of a projectile for which both the height reached and the distance traveled are defined. From this type of movement the lion reaches a height (H) of 3m and travels a horizontal distance (R) of 10 m. Mathematically the equations that describe this movement are given as,
Dividing the two equation we have that
Substituting values of H and R, we get
Substituting the value of \theta in equation we get,
Therefore the speed of the mountain lion just as it leaves the ground is 9.98m/s at an angle of 50.2°
