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Elodia [21]
3 years ago
11

Object C has a mass of 3,600 kilograms. Object D has a mass of 900 kilograms. Both objects were placed on different planets so t

hat their weights are equal. The gravity acting on Object C is _____ times the gravity acting on Object D.
Physics
2 answers:
Morgarella [4.7K]3 years ago
8 0

Answer:

1/4

Explanation:

The weight of an object is given by:

W=mg

where m is the mass of the object and g is the acceleration due to gravity on the planet.

We are told that object C and object D have same weight, so we can write:

m_C g_C = m_D g_D

where:

m_C = 3600 kg is the mass of object C

m_D = 900 kg is the mass of object D

We can re-arrange the equation to find a relationship between the gravity acting on object C and the gravity acting on object D:

g_C = \frac{m_D}{m_C}g_D = \frac{900 kg}{3600 kg}g_D = \frac{1}{4}g_D

So, the gravity acting on object C is 1/4 times the gravity acting on object D.

jek_recluse [69]3 years ago
7 0
The answer is 1/4
this is actually a very simple question because if you divide 3,600 by 4 that equals 900 so if you want them to be the same waight you need 3,600 to be multiplied by 1/4. 

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an apple falling off a tree

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The Atomic number tells us the number of ____ in an atom.
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Read 2 more answers
A simple pendulum is made by tying a 2.44 kg stone to a string 4.57 m long. The stone is projected perpendicularly to the string
alexdok [17]

Answer:

v_{max}=8.2226m/s

Explanation:

The problem is solved using the law of conservation of energy,

So

mgL(1-cos\theta)+\frac{1}{2}mv^2_0=\frac{1}{2}mv^2_{max}

v_{max}=\sqrt{2gL(1-cos\theta)+v^2_0}

v_{max}=\sqrt{2(9.8)(4.57)(1-cos(69.4))+8^2}

v_{max}=8.2226m/s

8 0
3 years ago
Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which
antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

5 0
2 years ago
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