Which of the following can be determined by studying the spectrum of a star?

A student practicing for a cross country meet runs 250 m in 30 s. What is her average speed?

Novay_Z [31]

It's 8.3 m per second

6 0

3 years ago

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

3 years ago

How much time does it take light from a flash camera to reach a subject 6.0 meters across a room in scientific notation?

wariber [46]

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

Explanation:

Given

t=?

d=6m

v=3*10^8 m/s

we have, v=d/t

here t=d/v

t=6m/3*10^8 m/s

v=2*10^-8 m/s

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

6 0

2 years ago

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

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;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

(c) From any of the equations, solve for Length of the string (L);

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$