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3 years ago

Which of the following is the same in all frames of reference?

1 answer:

4 0

The correct option is B.
The length of an object, the mass of an object and the rate of time passage for an object can change depending on the situation which the object is subject to. For instance in space, the mass and the velocity of an object usually change. But, the value of the speed of light in the space is the same for all observers regardless of the motion of an object, that is, the speed of light is a constant.<span />

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Briefly describe the characteristics of each soil horizon from the top layer to the bottom layer

inna [77]

<span>There is six horizen. 1. O Horizon - The top, organic layer of soil, 2. A Horizon - The layer called topsoil; 3. E Horizon - This layer is beneath the A Horizon and above the B Horizon. It is made up mostly of sand. 4. B Horizon - Also called the subsoil - this layer is beneath the E Horizon and above the C Horizon. 5. C Horizon - it's called regolith: the layer beneath the B Horizon and above the R Horizon. 6 R Horizon - this is last and the unweathered rock layer that is beneath all the other layers.</span>

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3 years ago

Which of the following should NOT be discussed during safety training

Juli2301 [7.4K]

Is there suppose to be an image?

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3 years ago

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

$\frac{1}{2}mv^2 = mgh$

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

3 0

3 years ago

What is the numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall?

Nuetrik [128]

Answer:

9.8 m/s/s

Explanation:

The numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall is 9.8 m/s/s. This is called the acceleration due to gravity.

8 0

3 years ago

Read 2 more answers

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

$T^2 = kR^3$

Substituting the values, we get the value of constant k for mars.

$687^2 = k\times (2.279 \times 10^{11})^3$

$k = 3.92 \times 10^{-29}$

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$

$R^3 = 3.39 \times 10^{33}$

$R= 1.50 \times 10^{11}\;\rm m$

Hence we can conclude that the distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

6 0

2 years ago