What term refers to a universal fact sometimes based on mathematical equations?

A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at ti

otez555 [7]

To solve the problem it is necessary to identify the equation in the manner given above.

This equation corresponds to the displacement of a body under the principle of simple harmonic movement.

Where,

$\xi = Acos(\omega t +\phi)$

PART A) Our equation corresponds to

$y = -5cos(4\pi t)$

Therefore the value of omega is equivalent to that of

$\omega = 4\pi$

From the definition we know that the period as a function of angular velocity is equivalent to

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{4\pi}$

$T = \frac{1}{2}$

This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the $cos\theta$ Is equivalent to . So the maximum speed that the body can reach is,

$y = -5cos(4\pi t)$

$y = -5cos(4\pi (1/2))$

$y = -5*(-1)$

$y = 5$

Therefore the maximum felocity will be 5ft / s

PART B) The period of graph is the time taken to reach from one maximum point to next point maximum point, then

$t = \frac{T}{2} = \frac{1}{2}*\frac{1}{2}$

$t = \frac{1}{4}s$

5 0

2 years ago

A woman gets burned at the beach after a hot, sunny day. What kind of

igomit [66]

Answer:

thermal, hot equals thermal and such

3 0

2 years ago

Read 2 more answers

The following questions consist of a statement and a reason .Answer these questions selecting the appropriate options given belo

Mamont248 [21]

Answer:

1)A)

1)A)2)A)

<h3>Hope it helps you ♡♡</h3>

4 0

2 years ago

A force of 6600 N is exerted on a piston that has an area of 0.010 m2

sveticcg [70]

Answer:

Choice A: approximately $0.015\; \rm m^2$ , assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

$\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}$ .

By Pascal's Principle, because the first piston exerted a pressure of $6.6\times 10^{5}\; \rm N \cdot m^{-2}$ on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be $6.6\times 10^{5}\; \rm N \cdot m^{-2}$ . In other words:

$P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}$ .