The relationship between the period of an oscillating spring and the attached mass determines the ratio of the period to 
.
Response:
- The ratio of the period to is always approximately<u> 2·π : 1</u>
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<h3>How is the value of the ratio of the period to
calculated?</h3>
Given:
The relationship between the period, <em>T</em>, the spring constant <em>k</em>, and the
mass attached to the spring <em>m</em> is presented as follows;
Therefore, the fraction of of the period to , is given as follows;
2·π ≈ 6.23
Therefore;
Which gives;
- The ratio of the period to is always approximately<u> 2·π : 1</u>
Learn more about the oscillations in spring here:
brainly.com/question/14510622
The possible direction of movement of an object accounts for the motion.
The cup shape of the socket allows for nearly circular motion. Hence, option (D) is correct.
What is a Motion?
When an object tends to move from one point to another in a given interval of time, then the object is said to be in motion.
Ball-and-socket joints possess a rounded, ball-like end of one bone fitting into a cup-like socket of another bone. This system allows the greatest range of circular motion, as all movement types are possible in all directions.
Thus, we can conclude that the cup shape of the socket allows for nearly circular motion. Hence, option (D) is correct.
Learn more about the motion of object here:
brainly.com/question/1065829
105/2 = 52.5/1
The rate of deceleration is 52.5 mps (miles per second).
Answer:
1.52 * 10⁵ K
Explanation:
When the temperature of a blackbody radiator increases, the overall radiated energy increases and the peak of the radiation curve moves to shorter wavelengths. When the maximum is evaluated from the Planck radiation formula, the product of the peak wavelength and the temperature is found to be a constant.
By Wien's Law,
λ * T = 2.898 * 10⁻³ mK
∴ T = 2.898 * 10⁻³/λ
Given λ = 19 nm = 19 * 10⁻⁹
T = 2.898 * 10⁻³ / 19 * 10⁻⁹
=1.52 * 10⁵ K
M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet
By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
= 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.
The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J
If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN
Answer:
(a) 3750 J
(b) 62.5 kN
