A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T
1 answer:
Answer:
0.266 m
Explanation:
Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,
P= mv where p is momentum, m is mass and v is the speed of an object. In this case
where sunscripts p and b represent putty and block respectively, c is common velocity.
Substituting the given values then
3*8=v(15+3)
V=24/18=1.33 m/s
The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy
where k is spring constant and x is the compression of spring. Substituting the given values then
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Answer:
(a) the fundamental frequency of this string is 65 Hz
(b) the harmonics of the given frequencies are third and fourth respectively.
(c) the length of the string is 2.74 m
Explanation:
Given;
mass density of the string, μ = 3 x 10⁻³ kg/m
tension of the string, T = 380 N
resonating frequencies, 195 Hz and 260 N
For the given resonant frequencies;
(c) From any of the equations, solve for Length of the string (L);
(a) the fundamental frequency is calculated as;
(b) harmonics of the given frequencies;
the first harmonic (n = 1) = f₀ = 65 Hz
the second harmonic (n = 2) = 2f₀ = 130 Hz
the third harmonic (n = 3) = 3f₀ = 195 Hz
the fourth harmonic (n = 4) = 4f₀ = 260 Hz
Thus, the harmonics of the given frequencies are third and fourth respectively.
Answer:
Centripetal acceleration = 83.77m/s²
Explanation:
<u>Given the following data;</u>
Radius, r = 0.13m
Velocity, v = 3.3m/s
To find centripetal acceleration;
Centripetal acceleration is given by the formula;
Substituting into the equation, we have;
<em>Centripetal acceleration = 83.77m/s²</em>
<em>Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s². </em>