Answer:
absorption and insolation.
Explanation:
Answer:
The horizontal distance travelled in that time lapse is 12.94 m
Explanation:
In order to solve this problem, we'll need:
- The horizontal speed
- the time the hammer takes to fall from the roof to the ground
At the lowest point of the roof, the hammer has a 9.88 m/s speed that makes an angle of 27° with the horizontal, so we can calculate the horizontal and vertical speed with trigonometry. If we take right as x positive and down as y positive we get

Now, we make two movement equation as we have a URM (no acceleration) in x and an ARM (gravity as acceleration) in y. We will wisely pick the lowest point of the roof as the origin of coordinates


Now we calculate the time the hammer takes to get to the floor
or 
Now, we keep the positive time result and calculate the horizontal distance travelled

(1) The wavelength of the wave is 1.164 m.
(2) The velocity of the wave is 23.7 m/s.
(3) The maximum speed in the y-direction of any piece of the string is 6.14 m/s.
<h3>
Wavelength of the wave</h3>
A general wave equation is given as;
y(x, t) = A sin(Kx - ωt)
<h3>Velocity of the wave</h3>
v = ω/K
From the given wave equation, we have,
y(x, t) = 0.048 sin(5.4x - 128t)
v = ω/K
where;
- ω corresponds to 128
- k corresponds to 5.4
v = 128/5.4
v = 23.7 m/s
<h3>Wavelength of the wave</h3>
λ = 2π/K
λ = (2π)/(5.4)
λ = 1.164 m
<h3>Maximum speed of the wave</h3>
v(max) = Aω
where;
- A is amplitude of the wave
- ω is angular speed of the wave
v(max) = (0.048)(128)
v(max) = 6.14 m/s
Thus, the wavelength of the wave is 1.164 m.
The velocity of the wave is 23.7 m/s.
The maximum speed in the y-direction of any piece of the string is 6.14 m/s.
Learn more about wavelength here: brainly.com/question/10728818
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To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.
The extension of the spring due to the weight of the object on Earth is 0.3m, then


The extension of the spring due to the weight of the object on Moon is a value of
, then

Recall that gravity on the moon is a sixth of Earth's gravity.




We have that the displacement at the earth was
, then


Therefore the displacement of the mass on the spring on Moon is 0.05m