What is most often given a value of zero to describe an object position on a srtaight line

Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a

____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, $v=2.79\times 10^8\ m/s$

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

$t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s$

Time period of oscillation measured by the observer is :

$t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s$

So, the time period of oscillation measured by the observer is 5.79 seconds.

3 0

2 years ago

A tire sits atop a ramp. When the tire is released, it rolls down the ramp. At the ramp's bottom, the speeding tire knocks down

k0ka [10]

Answer:

mechanical energy

Explanation:

Mechanical energy is the combination of both potential energy and kinetic

Mechanical energy can be divided as

1)kinetic energy, this energy vis regarded as the energy of motion

2) potential energy which is the stored energy of position.

Mechanical energy reffered to as

motion energy this energy is responsible for the movement of an object based on its position as well as motion.

Mechanical energy= U + K

Where U= potential energy

K= Kinectic energy

As the tire is sitting on top of a ramp, it posses "potential energy" as it is released and rolls down the ramp the potential is converted to Kinectic energy

4 0

2 years ago

In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar

katrin2010 [14]

Time period of any moon of Jupiter is given by

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

from above formula we can say that mass of Jupiter is given by

$M = \frac{4 \pi^2 r^3}{GT^2}$

now for part a)

$r = 4.22 * 10^8 m$

T = 1.77 day = 152928 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}$

$M = 1.9* 10^{27} kg$

Part B)

$r = 6.71 * 10^8 m$

T = 3.55 day = 306720 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}$

$M = 1.9* 10^{27} kg$

Part c)

$r = 10.7 * 10^8 m$

T = 7.16 day = 618624 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}$

$M = 1.89* 10^{27} kg$

PART D)

$r = 18.8 * 10^8 m$

T = 16.7 day = 1442880 seconds

now by above formula

$M = \frac{4 \pi^2 r^3}{GT^2}$

$M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}$

$M = 1.889* 10^{27} kg$

6 0

2 years ago

In what ways would organisms living in very deep water need to be specialized?

Nady [450]

Organisms living in great depths of water bodies like oceans and lakes need to be adapted for two (2) things especially; high water pressure and vision in darkness

The water column above from deep in the water can cause lots of hydrostatic pressure on the organisms’ cells. Also the fact that light cannot penetrate to great depth of water due to diffusion means the organisms must live in darkness.

Explanation:

It has been shown that cells from Piezophile bacteria have a high percentage of fatty acids in their membranes to prevent the cells from compacting solid from the high pressure.

Most of the organisms are also detritivores and use chemosynthesis for the autotrophs because light cannot reach these depths and hence photosynthesis is not possible. Organisms with eye vision are adapted to high wavelength light that can at least reach greater depths before diffusing. Nonetheless natural selection would favour use of sight for most organisms in this benthic region.

Learn More:

For more on adaptation check out;

brainly.com/question/12959056

brainly.com/question/350553

#LearnWithBrainly

8 0

3 years ago

Recently, astronomers have observed stars and other objects that orbit the center of the Milky Way Galaxy farther out than our S

Lesechka [4]

Answer:

That scenario can be explained by the idea of the contribution of dark matter on that point.

Explanation:

It can be explained through the idea of dark matter, this one was born to explain why stars (or any object) that were farther for the supermassive black hole in the center of the Milky Way galaxy didn't decrease it rotational velocity as it was expected according to equation 1.

$v = \sqrt{\frac{G M}{r}}$ (1)

Where v is the rotational velocity, G is the gravitational constant, M is the mass of the supermassive black hole, and r is the orbital radius.

Notice, that If the distance increases the orbital speed decreases (inversely proportional).

7 0

2 years ago