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Rainbow [258]
3 years ago
15

What are some things that doesn't move in a circular motion or are non examples of circular motion?

Physics
2 answers:
photoshop1234 [79]3 years ago
7 0

Answer:In non-uniform circular motion, an object's motion is along a circle, but the object's speed is not constant. In particular, the following will be true. The object's velocity vector is always tangent to the circle. The speed and angular speed of the object are not constant

Explanation:In non-uniform circular motion an object is moving in a circular path with a ... Since the speed is changing, there is tangential acceleration in ... Here is an example with an object traveling in a straight path

GuDViN [60]3 years ago
3 0
1: Радиус окружности постоянен (как при движении по круговой железной дороге или по автомобильной дороге). Изменение в
v
v
изменит величину радиального ускорения. Это означает, что центростремительное ускорение не является постоянным, как в случае с равномерным круговым движением. Чем больше скорость, тем больше радиальное ускорение. Частице, движущейся с более высокой скоростью, потребуется большая радиальная сила, чтобы изменить направление, и наоборот, когда радиус круговой траектории постоянен.

2: Радиальная (центростремительная) сила постоянна (как спутник, вращающийся вокруг Земли под действием постоянной силы тяжести). Круговое движение регулирует радиус в зависимости от изменения скорости. Это означает, что радиус круговой траектории является переменным, в отличие от случая равномерного кругового движения. В любом случае уравнение центростремительного ускорения в терминах «скорости» и «радиуса» должно выполняться. Здесь важно отметить, что, хотя изменение скорости частицы влияет на радиальное ускорение, на изменение скорости не влияет ни радиальная, ни центростремительная сила. Нам нужна тангенциальная сила, чтобы повлиять на изменение величины тангенциальной скорости. Соответствующее ускорение называется тангенциальным ускорением.
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fiasKO [112]
Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m
   = (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2
                       = (9.1×10∧-31 × 1,153,846.154∧2) ÷2
                      = (1.21154×10∧-18) ÷ 2
                      = 6.05769×10∧-19 J
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A train starts from rest and travels for 5.0 s with a uniform acceleration of 1.5 m/s2. What is the final velocity of the train?
alexandr1967 [171]

Answer:

Final speed of the train is 7.5 m/s

Explanation:

It is given that,

Uniform acceleration of the train is, a = 1.5 m/s²

It starts from rest and travels for 5.0 s. We have to find the final velocity of the train. By using first equation of motion as :

v=u+at

Here, train starts from rest so, u = 0

v=0+1.5\ m/s^2\times 5\ s  

v = 7.5 m/s

So, the final velocity of the train is 7.5 m/s. Hence, this is the required solution.

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3 years ago
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What is Newton's Second Law?
Rudiy27

Answer:

An object's acceleration depends on its mass and on the net force acting on it.

Explanation:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

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Please do number 25! Explain how you got your answer with detail to get Brainliest! Thank you!
Ad libitum [116K]
John weighs 200 pounds.
In order to lift himself up to a higher place, he has to exert force of 200 lbs.

The stairs to the balcony are 20-ft high.
In order to lift himself to the balcony, John has to do
(20 ft) x (200 pounds)  =  4,000 foot-pounds of work.

If he does it in 6.2 seconds, his RATE of doing work is
(4,000 foot-pounds) / (6.2 seconds)  =  645.2 foot-pounds per second.

The rate of doing work is called "power".

(If we were working in the metric system (with SI units),
the force would be in "newtons", the distance would be in "meters",
1 newton-meter of work would be 1 "joule" of work, and
1 joule of work per second would be 1 "watt".
Too bad we're not working with metric units.)

So back to our problem.

John has to do 4,000 foot-pounds of work to lift himself up to the balcony,
and he's able to do it at the rate of 645.2 foot-pounds per second.

Well, 550 foot-pounds per second is called 1 "horsepower".

So as John runs up the steps to the balcony, he's doing the work
at the rate of

           (645.2 foot-pounds/second) / (550 ft-lbs/sec per HP)

=  1.173 Horsepower.  GO JOHN !

(I'll betcha he needs a shower after he does THAT 3 times.)
_______________________________________________

Oh my gosh !  Look at #26 !  There are the metric units I was talking about.

Do you need #26 ?

I'll give you the answers, but I won't go through the explanation,
because I'm doing all this for only 5 points.

a).  5
b).  750 Joules
c).  800 Joules
d).  93.75%

You're welcome.

And #27 is 0.667 m/s .
7 0
3 years ago
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