Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
Friction<span> is the resistance to motion of one object moving relative to another. It is not a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.
Hope this is of great help to you, and happy studying~!
~Mistermistyeyed.</span>
The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s
Given parameters
To find
Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.
In this case we have a vertical launch
y = y₀ + v₀ t - ½ g t²
Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time
With the ball in hand, its position is zero
0 = 0 + v₀ t - ½ g t²
v₀ t - ½ g t² = 0
v₀ = ½ g t
Let's calculate
v₀ = ½ 9.8 2.4
v₀ = 11.76 m / s
In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s
Learn more about vertical launch kinematics here:
brainly.com/question/15068914
Answer:
Explanation:
given,
Wave vibrates = 37.6
time = 27.9 s
maximum distance travel = 450 cm
time = 11.3 s
wavelength = ?
frequency of wave
f = 1.35 Hz
Speed of wave
v = 39.82 cm/s
wavelength of wave
v = fλ
Hence, wavelength of the wave is equal to 25.79 cm.
