Wavelength, Wavespeed and frequency depends on each other in the following way :
And we know the Reciprocal relationship between frequency and period ~
So, let's find it's period •
A man jumps from a 45 m cliff into a large body of water. How long will he fall before he enters the wate
1 answer:
You might be interested in
We can use the law of conservation of energy to solve the problem.
The total mechanical energy of the system at any moment of the motion is:
where U is the potential energy and K the kinetic energy.
At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:
But the mechanical energy must be conserved, Ef=Ei, so we have
and so, the potential energy at the top of the flight is
The total mechanical energy of the system at any moment of the motion is:
where U is the potential energy and K the kinetic energy.
At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:
But the mechanical energy must be conserved, Ef=Ei, so we have
and so, the potential energy at the top of the flight is
Answer:
The tensions in is approximately 4,934.2 lb and the tension in
is approximately 6,035.7 lb
Explanation:
The given information are;
The angle formed by the two rope segments are;
The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°
The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°
Therefore, we have;
The tension in rope segment BC =
The tension in rope segment BD =
The tension in rope segment AB = = Pulling force of tugboat = 1200 lb
By resolution of forces acting along the line A_F gives;
× cos(26.0°) +
× cos(21.0°) =
= 1200 lb
× cos(26.0°) +
× cos(21.0°) = 1200 lb............(1)
Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;
× sin(26.0°) +
× sin(21.0°) = 0...........................(2)
Which gives;
× sin(26.0°) = -
× sin(21.0°)
= -
× sin(21.0°)/(sin(26.0°)) ≈ -
× 0.8175
Substituting the value of, , in equation (1), gives;
- × 0.8175 × cos(26.0°) +
× cos(21.0°) = 1200 lb
- × 0.7348 +
×0.9336 = 1200 lb
×0.1988 = 1200 lb
≈ 1200 lb/0.1988 = 6,035.6938 lb
≈ 6,035.6938 lb
≈ -
× 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb
≈ -4934.1733 lb
From which we have;
The tensions in ≈ -4934.2 lb and
≈ 6,035.7 lb.