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Ann [662]
2 years ago
7

Calculate the weight of a 2kg chicken on earth. .. (use the formulas)​

Physics
1 answer:
Aloiza [94]2 years ago
6 0

Answer:

weight =m x g =2kg × 10m/s 2 = 20N

<h3>Formulas:</h3>

weight = m × gravity

=unit for weight is newtons

Mass=weight ÷ gravity

= unit for mass is kilograms

Gravity=weight ÷ mass

=unit for gravity is 10m/sec 2

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In the water cycle, which state of matter has the particles closest together?
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An object is stopped at point ‘a’, and travels to point ‘c’ in 8s. What was its acceleration?
lutik1710 [3]

In SI units, its acceleration is  (the distance from A to C) / 32  m/s^2 .

7 0
3 years ago
A tomato of mass 0.18 kg is dropped from a tall bridge. If the tomato has a speed of 11 m/s just before it hits the ground, what
kondor19780726 [428]
The kinetic energy of the tomato is : 

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7 0
3 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
Radio waves transmitted through space at 3.00 ✕ 108 m/s by the Voyager spacecraft have a wavelength of 0.137 m. What is their fr
fredd [130]

Answer:

Frequency, f=2.18\times 10^9\ Hz

Explanation:

We have,

Speed of radio waves is 3\times 10^8\ m/s

Wavelength of radio waves is \lambda=0.137\ m

It is required to find the frequency of the radio waves. The speed of a wave is given by :

v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.137}\\\\f=2.18\times 10^9\ Hz

So, the frequency of the radio wave is 2.18\times 10^9\ Hz.

8 0
2 years ago
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