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matrenka [14]
3 years ago
5

Which type of motion did Aristotle associate the earth with

Physics
1 answer:
PtichkaEL [24]3 years ago
4 0

Answer: a circluar motion.

Explanation:

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Use F = 1/T as your basis:
skelet666 [1.2K]
  • Time Period=T=2.5×10^{-3}s

Now

Frequency:-

\\ \rm\rightarrowtail \nu=\dfrac{1}{T}

\\ \rm\rightarrowtail \nu=\dfrac{1}{2.5\times 10^{-3}}

\\ \rm\rightarrowtail \nu=0.4\times 10^{3}

\\ \rm\rightarrowtail \nu=400Hz

6 0
2 years ago
At what speed, as a fraction of c, does a moving clock tick at four fifth the rate of an identical clock at rest?
ololo11 [35]

Answer:

The seed as a fraction of the speed of light is \frac{3}{5}c

Solution:

As per the question:

Suppose, t_{i} be the rate of an identical clock between two time intervals.

For a moving clock, moving with velocity 'v', at the clock tick of four-fifth:

t = \frac{5}{4}t_{i}

Now,

Using the relation of time dilation, from Einstein's relation:

t = \frac{t_{i}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

\frac{5}{4}t_{i} = \frac{t_{i}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

Squaring both sides:

(\frac{5}{4})^{2} = (\frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}})^{2}

\frac{25}{16} = \frac{1}{{1 - \frac{v^{2}}{c^{2}}}}

1 - \frac{16}{25} = \frac{v^{2}}{c^{2}}

\frac{v}{c} = \sqrt{\frac{9}{25}}

\frac{v}{c} = \frac{3}{5}

v = \frac{3}{5}c

6 0
3 years ago
PLS HELP!! YOU CAN SKIP THE INFO IF YOU WANT!!
kotegsom [21]

Answer:b,c,d I think

Explanation:

B because it moves sediment, or sand, to the ocean, c because it pollutes the ocean with plastic and stuff, and d because the bacteria can be harmful

8 0
3 years ago
Read 2 more answers
A beach ball moving with a speed of 1.31 m/s rolls off a pier and hits the water 0.88 m from the end of the pier
slavikrds [6]
So what is the main question? I Need to know what the main question is to solve it.
5 0
4 years ago
Please show work : A particle with mass 2.00 μg and a charge of – 200 nC has a velocity of 3000 m/s in the x-direction. There is
irga5000 [103]

Answer:

 x =4.5 10⁴ m

Explanation:

To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law

     m = 2.00 mg (1 g / 1000 ug) (1 Kg / 1000g) = 2.00 10-6 Kg

     q = -200 nc (1C / 10 9 nC) = -200 10-9 C

Let's calculate the acceleration

     F = ma

     F = q E

     a = qE / m

     a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶

     a = 1 10² m / s²

Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)

     Vf² = Vo² -2 a x

     0 = Vo² - 2 a x

     x = Vo² / 2a

     x = 3000²/ 2100

     x =4.5 10⁴ m

This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial

Second part

In this case Newton's second law is applied on the y axis

      F -W = 0

      F = w = mg

      E q = mg

      E = mg / q

      E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹

      E = 9.8 10⁵ C

       

The direction of the field is such that the force on the particle is up, as the particle has a negative charge, the field must be directed downwards F = qE = (-q) E

7 0
4 years ago
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