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1 year ago

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calculate the weight of an object of mass 100 kg on the surface of a Planet whose mass and diameter are 4.8*10^24 kg and 12000km

respectively.

1 answer:

PilotLPTM [1.2K]1 year ago

6 0

The weight of an object mass of 100 kg on the surface of a Planet willl is 981 N.

<h3>What is weight?</h3>

The weight of matter is found as the product of the mass and the gravitational acceleration;

Given data;

Weight of an object,W =?

Mass,m = 100 kg

$\rm W= mg \\\\ \rm W= 100 \ kg \times 9.81 \ m/sec^2 \\\\ W=981 \ N$

Hence the weight of an object willl be 981 N.

To learn more about the weight refer;

brainly.com/question/10069252

#SPJ1

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2 years ago

Many minerals, such as silver, are good conductors of heat and electricity. How could technology use this property to help ident

Aleks [24]

Answer:

This property could be used to create technologically-advanced tools or machines that could easily locate the mineral deposits.

Explanation:

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If this technology will be implemented, though, regulation policy must be strictly implemented because it might lead to<em> over-mining</em> thus leading to the depletion of mineral deposits.

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3 years ago

Give two environmental effects of using wood as an energy resource.

vladimir2022 [97]

Answer:

Using wood also helps keep carbon out of the atmosphere, helping to mitigate climate change.

Deforestation.

Explanation:

Trees store carbon dioxide as they grow. After harvest, wood products continue to store much of this carbon. These benefits continue when wood is reclaimed to manufacture other products.

Extraction of wood within the supply areas clearly affects the forest and the environment. Some of these impacts can be seen in the study areas. The rainfall is increasingly irregular, biodiversity has been lost. Communities have experienced flooding and drought, which adversely affects the most important sector in the country's economy, that of agriculture. The local populations are concerned about these impacts, which affect not only the environment but the whole social economy.

6 0

2 years ago

Read 2 more answers

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

6 0

3 years ago

Please someone help, I’m very confused and it’s due soon, thanks

Anit [1.1K]

Answer:

1 s
19.6 m
2 s
0.8 m/s^2
28 m/s
79 m/s
0.37 s
26 m/s
242 m/s
19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

PE = Mgh

At velocity v, the kinetic energy of the object is ...

KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

PE = KE

Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

2gh = v^2

This can be solved for height:

h = v^2/(2g) . . . . [eq1]

or for velocity:

v = √(2gh) . . . . [eq2]

__

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

t = 2(v^2/(2g))/v = v/g . . . . [eq3]

t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

g = 2h/t^2 . . . . [eq6]

__

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

__

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

__

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

__

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

__

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

__

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

__

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

2(9/49 s) ≈ 0.37 s

__

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

__

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

__

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

<em>Additional comment</em>

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

5 0

3 years ago