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Sir Isaac Newton deduced that the weight of an object is due to the force of gravity - Leaving Cert Physics - Question 6 - 2014

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Sir Isaac Newton deduced that the weight of an object is due to the force of gravity. Define force and give the unit of force. State Newton's law of universal gravi... show full transcript

Worked Solution & Example Answer:Sir Isaac Newton deduced that the weight of an object is due to the force of gravity - Leaving Cert Physics - Question 6 - 2014

Step 1

Define force and give the unit of force.

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Answer

Force can be defined as a cause that changes the motion of an object. Mathematically, it is represented by the formula:

F=ΔpΔtF = \frac{\Delta p}{\Delta t}

where Δp\, \Delta p is the change in momentum, and Δt\, \Delta t is the change in time.

The unit of force is newton (N), which can also be expressed as kg m/s².

Step 2

State Newton's law of universal gravitation.

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Answer

Newton's law of universal gravitation states that the force of attraction between two masses is proportional to the product of the masses and inversely proportional to the square of the distance between their centers. This can be mathematically expressed as:

Fm1m2r2F \propto \frac{m_1 m_2}{r^2}

where m1m_1 and m2m_2 are the two masses, and rr is the distance between their centers.

Step 3

Use the equation below, … acceleration due to gravity on Mars.

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Answer

To find the acceleration due to gravity on Mars, we use:

g=GMr2g = \frac{GM}{r^2} where:

  • G=6.67×1011  N m2/kg2G = 6.67 \times 10^{-11} \; \text{N m}^2/\text{kg}^2,
  • M=6.4×1023  kgM = 6.4 \times 10^{23} \; \text{kg}, and
  • r=3.4×106  mr = 3.4 \times 10^6 \; \text{m}.

Substituting the values, we get:

g=(6.67×1011)(6.4×1023)(3.4×106)23.7  m/s2g = \frac{(6.67 \times 10^{-11}) (6.4 \times 10^{23})}{(3.4 \times 10^6)^2} \approx 3.7 \; \text{m/s}^2

Step 4

Give a reason for … rover was to be used on Earth.

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Answer

The wheels of the Curiosity rover are designed for the lower gravity on Mars. Since Mars has less gravitational force than Earth, the wheels would need to withstand different physical demands on Earth, which could lead to structural differences and make them less effective on solid surfaces.

Step 5

(i) the weight of Curiosity on Earth

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Answer

To find the weight of Curiosity on Earth, we use the formula:

W=mgW = mg where:

  • m=899  kgm = 899 \; \text{kg},
  • g=9.8  m/s2g = 9.8 \; \text{m/s}^2.

Thus,

W=899×9.88803  NW = 899 \times 9.8 \approx 8803 \; \text{N}

Step 6

(ii) the mass of Curiosity on Mars

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Answer

The mass of an object remains constant regardless of location. Therefore, the mass of Curiosity on Mars is still:

m=899  kgm = 899 \; \text{kg}

Step 7

(iii) the weight of Curiosity on Mars.

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Answer

To find the weight of Curiosity on Mars, we again use the weight formula:

W=mgW = mg where now:

  • g=3.7  m/s2g = 3.7 \; \text{m/s}^2.

Thus,

W=899×3.73326.3  NW = 899 \times 3.7 \approx 3326.3 \; \text{N}

Step 8

Give one other part of the electromagnetic spectrum.

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Answer

One other part of the electromagnetic spectrum is ultraviolet radiation.

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