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3 years ago

How do you find circular motion?

Physics

2 answers:

Ksivusya [100]3 years ago

7 0

Answer:

2πr

as we know that the circumference of a circle is 2 * π * r, the arc length of the whole circle is 2πr units. Then using the formula s = r * theta, the angle theta equals the arc length divided by the radius. Therefore theta = 2πr/r = 2π radians.

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MatroZZZ [7]3 years ago

6 0

Answer:

Explanation:

By multiplying the rotational frequency with the circumference we can determine the average speed of the object. The circular velocity formula is expressed as, vc = 2 πr / T. Where in, r denotes the radius of the circular orbit. T is time period.

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What are two main types of friction

emmainna [20.7K]

Answer:There are two main types of friction, static friction and kinetic friction. Static friction operates between two surfaces that aren't moving relative to each other, while kinetic friction acts between objects in motion.

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3 years ago

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Discuss Joule-Thompson effect with relevant examples and formulae.

Delicious77 [7]

Answer:

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

Explanation:

Joule -Thompson effect

Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process

Let

$P_1,T_1$ Pressure and temperature at inlet and

$P_2,T_2$ Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

$\mu _j=\left(\frac{\partial T}{\partial p}\right)_h$

Now from T-ds equation

dh=Tds=vdp

$Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp$

⇒ $dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

So Joule -Thompson coefficient

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

$\dfrac{\partial v}{\partial T}=\dfrac{v}{T}$

So by putting the values in

$\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp$

$\mu _j=0$ For ideal gas.

6 0

3 years ago

Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mas

timurjin [86]

Answer:

0.247 J = 247 mJ

Explanation:

From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)

= 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)

= 0.135 J + 0.11220 J

= 0.2472 J

≅ 0.247 J = 247 mJ

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3 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

valentina_108 [34]

Answer:

-time it takes for the sled to come to a stop after launch of rocket = 7.244 s

-distance sled has travelled from its starting point by the time it finally comes to rest is = 234.8655 m

Explanation:

From the question, looking at the motion while accelerating, we have;

Initial velocity; u = 0 m/s

Acceleration; a = 13.5 m/s²

Time; t = 3.3 s

Let's use first equation of motion to find final velocity (v).

v = u + at

v = 0 + (13.5 × 3.3)

v = 44.55 m/s

In this forward direction, let's calculate the displacement(d1) using newton's 3rd equation of motion.

d1 = ut + ½at²

d1 = 0(3.3) + ½(13.5 × 3.3²)

d1 = 73.5075 m

Now, let's consider the motion while slowing down and our final velocity will be 0 m/s while initial velocity will now be 44.55 m/s while acceleration is 6.15 m/s².

Thus, from v = u + at, we can find the time it take for the sled to come to a stop.

Now, since it's coming to rest acceleration will be negative. Thus;

0 = 44.55 + (-6.15t)

0 = 44.55 - 6.15t

t = 44.55/6.15

t = 7.244 s

Now we want to find out how far the sled has travelled from its starting point by the time it finally comes to rest.

Thus, we'll use the equation;

v² = u² + 2as

Where s will be the second displacement which we will call d2.

Thus;

0² = 44.55² + (-2 × 6.15 × s)

0 = 1984.7025 - 12.3s

12.3s = 1984.7025

s = 1984.7025/12.3

s = 161.358

Thus, d2 = s = 161.358 m

Thus, distance sled has travelled from its starting point by the time it finally comes to rest is ;

= d1 + d2 = 73.5075 + 161.358 = 234.8655 m

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4 years ago

The existence of which aspect of the universe supports the big bang theory? cosmic background radiation galactic clusters irregu

zysi [14]

The big bang theory refers to a theoretical concept that explains the creation of our universe. It postulates that the universe started with a giant explosion that created all matter and propelled it outwards in all directions. It is the existence of cosmic background radiation that supports the big bang theory within cosmology.

7 0

3 years ago

Read 2 more answers

How do you find circular motion?​

How do you find circular motion?