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laiz [17]
3 years ago
10

Gravity on earth is 9.8 m/s squared, and gravity on the moon is 1.6 m/s squared. So if the mass of an object on earth is 40 kilo

grams, its mass on the moon is
Physics
1 answer:
garik1379 [7]3 years ago
7 0

The mass of an object on Earth is the same as its mass on the Moon. The weight is different.

Weight = m * g

Weight ( Moon ) = 40 kg * 1.6 m/s² = 64 N

If the mass of an object on Earth is 40 kg, its mass on the Moon is 40 kg and its weight on the Moon is 64 N. 

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An asteroid orbits the Sun every 176 years. What is the asteroids average distance from the Sun? P ^ 2 = a ^ 3 where p = period
KatRina [158]

Answer:

The value is  x =  45.99 \  Au

Explanation:

From the question we are told that

   The period of the  asteroid is   T =  176 \ years = 176 * 365 * 24 * 60* 60 = 5.55*10^{9}\ s

Generally the average distance of the asteroid from the sun is mathematically represented as

            R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }

Here M is the mass of the sun with a value  

        M  =  1.99*10^{30} \  kg

         G  is the gravitational constant with value  G  =  6.67 *10^{-11}  \  m^3 \cdot kg^{-1} \cdot  s^{-2}

           R = \sqrt[3]{ \frac{6.67 *10^{-11}  * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }

=>       R = 6.88 *10^{12} \  m

Generally

         1.496* 10^{11}  \  m  \to  1 Au (Astronomical \  unit )

So

          R = 6.88 *10^{12} \  m \ \ \ \ \to \ \   x \  Au

=>      x =  \frac{6.88 *10^{12}}{1.496 *10^{11}}

=>       x =  45.99 \  Au

       

7 0
2 years ago
How many times can mechanical energy become chemical energy
maria [59]

Answer:

Chemical energy, Energy stored in the bonds of chemical compounds. Chemical energy may be released during a chemical reaction, often in the form of heat; such reactions are called exothermic.

3 0
3 years ago
You need to build a prototype with machined parts that withstand a saline corrosive environment and temperatures above 200 degre
kondor19780726 [428]

Answer:

Stainless steel

Explanation:

I will try to order the solutions from the least correct to the most correct.

Since a temperature greater than 200 ° F is required, that is to say approximately 93 ° c, <em>Polycaprolactone</em> is the least indicated. Its melting point is approximately 60 ° C, so it would not serve the required application.

On the other hand we have<em> Untreated aluminum</em>, which although it has a melting point higher than the required one, without a zinc and magnesium treatment it will easily oxidize in a salty environment, so it cannot be used in this choice either.

We have to compare the two steels.

The<em> Mild Steel </em>has a better corrosion resistance than the previous ones, but in a long-term cycle it will end up full of corrosion and therefore its properties will be highly affected.

Finally, we have <em>stainless steel</em>, which, as the name implies, contains in some of its variations chromium, zinc or magnesium in its alloys, which makes it highly resistant to corrosion.

In addition its melting point is above 1500 ° c.

The best choice is stainless steel.

5 0
3 years ago
TC
frosja888 [35]

Answer:

0.08 sin 3nt +

metre. Then calculate-

a time period

(d) Time period

(

Minitial phase

wõisplacement

7

(c) Displacement from mean position at t=

36

sec.

Explanation:

thats the answer

6 0
3 years ago
Spiral fracture of bone: Spiral fracture of bone occurs due to twisting of the limb, and is a very common skiing accident. The f
Pavlova-9 [17]

Answer:

principal stresses :б1 = 32.62mPa  б2 = 31.38mPa

Max Shear stress : 16.31 mPa

Orientation of max principle plane = 44.43°

Orientation of minimum principal plane = 134.43°

Explanation:

Given data:

Torque = 50 N-m

weight = 80 kgs

half of weight is subjected to each leg

radius of bone = 10 mm = 0.010 m

<u>a) Determine the principal stresses and shear stress</u>

first calculate the max shear stress ( this will occur in the outermost element

= 16T / π*d^3   where : T = 50 , d = 0.020 m

hence max shear stress = 32 mPa

next determine compressive stress

= ( 40*g)  / π/4*d^2 . where : d = 0.020 m , g = 9.81

hence compressive stress = 1.24 mPa

draw and calculate the radius of Mohr's circle

radius of Mohr's circle = 32.0060

Hence principal stresses = 32.0060 ± 0.62

б1 = 32.62mPa  

б2 = 31.38mPa

attached below is the remaining part of the solution

3 0
3 years ago
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