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slavikrds [6]
3 years ago
9

Atoms with atomic number ____ or less will not undergo fission.

Physics
1 answer:
BlackZzzverrR [31]3 years ago
7 0
26

I hope its helped you a lot good luck
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1. George is traveling to Boston from Springfield. Springfield is 144
denpristay [2]

Answer:

48 kilometers per hour

Explanation:

144/3 = 48 km/h

3 0
2 years ago
A dog of mass 4 kg runs up a hill of height 8 m. How much gravitational potential energy does the dog gain?
Genrish500 [490]
A. 314 because when you use the formula for the GPE ; GPE=MGH or means mass times gravity time height (4x8x9.8) and thats equivalent to 313.6 which rounds up to 314. Hope it helps 
8 0
3 years ago
Read 2 more answers
An object travels with a constant speed in a circular path. The net force on the object is
Pepsi [2]

Answer:

toward the center

Explanation:

Before answering, let's remind the first two Newton Laws:

1) An object at rest tends to stay at rest and an object moving at constant velocity tends to continue its motion at constant velocity, unless acted upon a net force

2) An object acted upon a net force F experiences an acceleration a according to the equation

F=ma

where m is the mass of the object.

In this problem, we have an object travelling at constant speed in a circular path. The fact that the trajectory of the object is circular means that the direction of motion of the object is constantly changing: this means that its velocity is changing, so it has an acceleration. And therefore, a net force is acting on it. The force that keeps the object travelling in the circular path is called centripetal force, and it is directed towards the center of the circle (because it prevents the object from continuing its motion straight away).

So, the correct answer is

toward the center

8 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
Period T= time of one revolution measured in seconds
ser-zykov [4K]
60 RPM equals one hertz (i.e., one revolution per second, or a period of one second). The SI unit for period is the second.
3 0
3 years ago
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