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Damm [24]
3 years ago
9

How big is a ton? That is, what is the volume of something that weighs a ton? To be specific, estimate the diameter of a 1-ton r

ock, but first make a wild guess>: will it be 1 ft across, 3 feet, or the size of a car? (Hint: Rock has mass er volume about three times that of water, which is 1 kg per liter(10^3cm^3) or 62lb per cubic foot) ...?
Physics
1 answer:
Zepler [3.9K]3 years ago
4 0
<span>Short ton is equal to 2000 lb. A metric tonne is equal to 1000 kilograms.

T</span>o get the volume of a rock you should divide the mass (1000 kg) by the density. Assume the density is three times of water. You can assume the rock is a sphere or cube to get the approximate dimension from the volume. 

<span>The density of water is 1000 kg / m^3
</span>
Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0
4 years ago
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
Steve does 80 J of work to a rope threaded through a pulley and attached to a crate of anvils. The tension of the rope is 10 N a
Ivahew [28]

Answer:

The Answer is (B)

Explanation:

6 0
3 years ago
Read 2 more answers
A cannon fires a shell straight upward; 2.3 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming
PtichkaEL [24]

Answer:

The speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.                    

Explanation:

Let u is the initial speed of the launch. Using first equation of motion as :

u=v-at

a=-g

u=v+gt\\\\u=17+9.8\times 2.3\\\\u=39.54\ m/s

The velocity of the shell at launch and 5.4 s after the launch is given by :

v=u-gt\\\\v=39.54-9.8\times 5.4\\\\v=-13.38\ m/s

So, the speed of the shell at launch and 5.4 s after the launch is 13.38 m/s it is moving towards the Earth.

6 0
3 years ago
The angle of incidence always equals what?
Ratling [72]

Answer:

Angle Of Incidence Formula The angle of incidence is equal to the reflected angle through the law of reflection. The angle of incidence and the angle of reflection is always equal, and they are both on the same plane along with the normal. 2,44,451

Explanation:

that could help you with work.

4 0
3 years ago
Read 2 more answers
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