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3 years ago

A toy airplane, flying in a horizontal, circular

Physics

2 answers:

Dmitriy789 [7]3 years ago

6 0

The average speed of the airplane is 8.4 m/s

<h3>Further Explanation</h3>

<u>Given :</u>

number of rotation = N = 10

time intervals = Δt = 30 s

radius = R = 4.0 m

<u>Unknown :</u>

average speed = v = ?

<u>Solution :</u>

v = ω × R

v = ( θ / Δt ) × R

v = ( 10 × 2π / 30 ) × 4

v = (8π / 3) m/s

<h3>Learn more</h3>

Newton's Law of Motion: brainly.com/question/10431582

Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Newton, Law, Impulse, Work

Goshia [24]3 years ago

4 0

Answer:

8.4 m/s

Explanation:

The toy completes 10 circle in 30 seconds. So its frequency of revolution is

$f=\frac{10}{30 s}=0.33 Hz$

The periof of revolution is the reciprocal of the frequency, so

$T=\frac{1}{f}=\frac{1}{0.33 Hz}=3 s$

The radius of the circular path is

r = 4.0 m

So the total distance covered by the toy in one circle is the length of the circumference:

$2\pi r$

And so the average speed is

$v=\frac{2\pi r}{T}=\frac{2\pi (4.0 m)}{3 s}=8.4 m/s$

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3 years ago

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Joe balances a stationary coin on the the tip of his finger 20 cm from the top of the table. How much work is Joe doing?

adell [148]

The work done by Joe is 0 J.

<u>Explanation</u>:

When a force is applied to an object, there will be a movement because of the applied force to a certain distance. This transfer of energy when a force is applied to an object that tends to move the object is known as work done.

The energy is transferred from one state to another and the stored energy is equal to the work done.

W = F . D

where F represents the force in newton,

D represents the distance or displacement of an object.

Force = 0 N, D = 20 cm = 0.20 m

W = 0 $\times$ 0.20 = 0 J.

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3 years ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

<u>For process 1:</u>

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

<u>For process 2:</u>

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

<u>For process 3:</u>

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

<u>For process 4:</u>

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

<u>For process 5:</u>

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

4 0

3 years ago

A solid cylinder of uniform density of 0.85 g/cm3 floats in a glass of water tinted light blue by food coloring.

SVEN [57.7K]

Each side has to have at least 44 horses

F61160 N. This is further explained below.

<h3>What is the force?</h3>

Generally, We are only interested in the component that operates horizontally since the vertical components all cancel each other out. The pressure difference works on the hemisphere to generate a normal force all over the surface, but we are only concerned with that force's horizontal component. This may be determined by supposing the hemispheres to be two flat circular plates of the same radius as the hemispheres that have been forced together.

Therefore, force is equal to pressure multiplied by area, which is

F= (970 -15 )( * (0.45 m)2)

F=60754 N for each side.

Therefore, each side has to have at least 44 horses

44* 1390 = 61160 N