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ivolga24 [154]
3 years ago
10

Which best explains the relationship between heat energy and temperature?

Physics
1 answer:
Vanyuwa [196]3 years ago
5 0
Heat energy is to a single organism
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A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s . A second block, sliding at a faster 4.0 m/sm/s , collides w
aleksklad [387]

Answer:

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block= ?

v₁= velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃= final velocity of combined block=1.8m/s

applying the formula above

(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

4m₂ - 1.8m₂ = 3.96 - 2.64

2.2m₂=1.32

divide both sides by 2.2

m₂= 0.6kg

4 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
When the velocity of an object Changes it is acted upon by a
anygoal [31]

When the velocity of an object changes, it is acted upon by a force

5 0
3 years ago
Determine la fuerza aplicada sobre un émbolo en una prensa hidráulica si se quiere cargar una lanza de 2970kg; toma en cuenta qu
Artist 52 [7]
I’m not sure sorry I really wish I could help
4 0
2 years ago
A string that is 3.6 m long is tied between two posts and plucked. The string produces a wave that has a frequency of 320 Hz and
marin [14]

To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as

\lambda = \frac{v}{f}

Where,

v = Velocity

f = Frequency,

Our values are given as

L = 3.6m

v= 192m/s

f= 320Hz

Replacing we have that

\lambda = \frac{192}{320}

\lambda = 0.6m

The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,

N = \frac{L}{\lambda}

N = \frac{3.6}{0.6}

N = 6

Therefore the number of wavelengths of the wave fit on the string is 6.

5 0
2 years ago
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