Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;

(a) the force constant of the spring

(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k

Wavelength = (speed) / (frequency)
= (3 x 10⁸ m/s) / (1 x 10⁸ /s) = 3 meters
Answer:
A. Measurements are made with tools; observations are not
Explanation:
Measurement: <em>the assignment of numbers or codes according to prior-set rules. </em>
Observation: <em>data from an individual study subject or sampled unit.</em>
Measurement error: <em>differences between "true" answers and what appears on data collection instruments</em>
hope this helps
plz mark brainleist
Answer:
reflection of water ,sound and water waves etc
Answer:
The maximum speed at which the car can safety travel around the track is 18.6m/s.
Explanation:
Since the car is in circular motion, there has to be a centripetal force
. In this case, the only force that applies for that is the static frictional force
between the tires and the track. Then, we can write that:

And since
and
, we have:

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

Substituting this expression for
and solving for
, we get:

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

It means that in its maximum value, the speed of the car is equal to 18.6m/s.