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NemiM [27]
3 years ago
5

The mass of an object is 84 kg. Find its weight on the Earth and the Moon.​

Physics
2 answers:
lozanna [386]3 years ago
8 0

Answer:

We would take the mass given, 80 kg, and multiply it times the acceleration of Earth's downward pull on that mass, 9.8 m/s/s. So the weight on Earth is 784 newtons. So the weight of the same 80 kg mass on the moon is 133.28 N.

zimovet [89]3 years ago
6 0

hi

mass is express  in Newtons.

On Earth gravity is  9.807   so  wieght would be :   9.807 * 84= 823.79

On the Moon garvity is  1.62  so weight would be  :  84 *1.62 = 136.08

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A force of 333 N is applied 33 cm from the pivot point. What is the maximum torque of this situation
exis [7]

Answer: 109.89 Nm

Explanation:

The maximum torque will be calculated as the force multiplied by the perpendicular distance. This will be:

Torque = force × perpendicular distance

torque = 333 × 0.33

= 109.89 Nm

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3 years ago
How much work is done when 0.0050 c is moved through a potential difference of 9.0 v? use w = qv?
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Answer:

0.045 J

Explanation:

The work done on a charge moving through a potential difference is given by

W=q\Delta V

where

W is the work done

q is the charge

\Delta V is the potential difference

In this problem, we have

q = 0.0050 C is the charge

\Delta V=9.0 V is the potential difference

Using the formula, we find the work done:

W=(0.0050 C)(9.0 V)=0.045 J

4 0
3 years ago
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URGENT! 50pts! Help with rotational speed questions please?
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3 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

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