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MrRissso [65]
3 years ago
8

Explain how evidence of energy transformation in fireworks supports the law of conservation of energy.

Physics
1 answer:
Varvara68 [4.7K]3 years ago
7 0

Energy transformation in fireworks supports the law of conservation of energy because the total chemical energy packed into the fireworks before they ignite must be the same as the total remaining after it explodes... I hope this was the answer you were looking for.. I love your Yoongi pfp btw! :) </3

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Observations must be _____. <br> subjective <br> objective <br> biased <br> deductive
Dmitrij [34]
Objective
because it means<span> based on measurement, or reasoning free of bias.</span>
6 0
3 years ago
Read 2 more answers
1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as
telo118 [61]

Explanation:

The tangential speed of Andrea is given by :

v=r\omega

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v

New angular speed is twice that of the Chuck's speed.

8 0
3 years ago
How can an object overcome static friction?
larisa86 [58]

Answer:

Applying enough force in one direction to move the object, making kinetic energy.

Explanation:

Simpleness

4 0
3 years ago
Future space stations could create an artificial gravity by rotating. Consider a cylindrical space station that rotates with a p
aleksandrvk [35]

Answer:

P = 2 pi R / v    period of space station

F / m = v^2 / R    centripetal force per unit of mass

So F / m = 4 pi^2 R^2 / (P^2 * R) = 4 pi^2 R / P^2

Also, F / m = 9.8 m/s^2   earth's gravitational attraction

So 9.8 = 4 pi^2 R / P^2    or    R = 9.8 P^2 / 4 * pi^2) = 195 m

Or D = 2 R = 390 m the diameter required

8 0
2 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
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