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Semenov [28]
2 years ago
11

g A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 ra

d/s. What is the constant angular acceleration (in rad/s) of the wheel
Physics
1 answer:
Setler [38]2 years ago
5 0

Answer:

The angular acceleration of the wheel is 15.21 rad/s².

Explanation:

Given that,

Time = 5 sec

Final angular velocity = 96.0 rad/s

Angular displacement = 28.0 rev = 175.84 rad

Let \alpha be the angular acceleration

We need to calculate the angular acceleration

Using equation of motion

\theta=\omega_{i} t+\dfrac{1}{2}\alpha t^2

Put the value in the equation

175.84=\omega_{i}\times 5+\dfrac{1}{2}\times\alpha\times(5)^2

175.84=\omega_{i}\times 5+12.5\alpha......(I)

Again using equation of motion

\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

96.0=\omega_{i}+\alpha \times 5

On multiply by 5 in both sides

480=\omega_{i}\times 5+\alpha\times 25....(II)

On subtract equation (I) from equation (II)

480-175.84=\alpha(25-5)

304.16=\alpha\times20

\alpha=\dfrac{304.16}{20}

\alpha=15.21\ rad/s^2

Hence, The angular acceleration of the wheel is 15.21 rad/s².

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Un recipiente contiene 224 dm3 de Ozono de masa 4.561 Kg a 51.09 grados celsius. Calcula la presión del Ozono
kari74 [83]

Answer:

Por lo tanto, la presión del ozono es:

P=0.011\: atm  

Explanation:

Podemos usar la ecuacion de los gases ideales;

PV=nRT (1)

Tenemos:

El volumen V = 224 dm³ = 224 L

La temperatura T = 51.09 C = 324.09 K

La masa es m = 4.561 kg

Lo necesitamos ahora es calvular n que es el numero de moles;

recordemos que el peso molecular del ozono M = 48 g/mol.

n=\frac{m}{M}=\frac{4.561}{48}=0.095\: mol

Finalmente, usando la ecuacion 1 despejamos la presion P

P=\frac{nRT}{V}

P=\frac{0.095*0.082*324.09}{224}  

Por lo tanto, la presion del ozono es:

P=0.011\: atm  

Espero te haya ayudado!

5 0
3 years ago
These plans are an accompaniment to mechanical lifting devices in reducing on-the-job injuries. they address specific patient ne
lilavasa [31]
For the answer to the question, ergonomics<span> is the science of fitting the job to the worker. When there is a mismatch between the physical requirements of the job and the physical capacity of the worker, work-related musculoskeletal disorders (MSDs) can result. Ergonomics is the practice of designing equipment and work tasks to conform to the capability of the worker, it provides a means for adjusting the work environment and work practices to prevent injuries before they occur.</span>
4 0
3 years ago
5. Kelvin applied a rightward force of 302 N to a 29 kg desk to accelerate it across the floor. The
Naddik [55]

The force of friction is 213.2 N

Explanation:

The frictional force acting on an object sliding on a surface is given by:

F_f = \mu mg

where:

\mu is the coefficient of friction

m is the mass of the object

g is the acceleration of gravity

In this problem we have

\mu=0.750

m = 29 kg

g=9.8 m/s^2

Therefore, the force of friction is

F_f = (0.750)(29)(9.8)=213.2 N

Learn more about friction:

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3 0
3 years ago
What would be the magnitude of the electric field 0.75 m from a 0.63 C master charge and what would be the force on a 0.50 C tes
bagirrra123 [75]

The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

<h3>Electric field on the master charge</h3>

E = kq/r²

where;

  • q is magnitude of master charge
  • r is distance of separation
  • k is Coulomb's constant

E = (9 x 10⁹ x 0.63)/(0.75²)

E = 1.008 x 10¹⁰ N/C

<h3>Force on the test charge</h3>

F = Eq

where;

  • E is electric field
  • q is the test charge

F = (1.008 x 10¹⁰) x (0.5)

F = 5.04 x 10⁹ N

Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

Learn more about electric field here: brainly.com/question/14372859

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8 0
2 years ago
An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-
maw [93]

Answer:

  1. Power requirement <u>P</u> for the banner is found to be  30.62 W
  2. Power requirement <u>P</u> for the solid flat plate is found to be 653.225 W
  3. Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The <em>Cd </em>for banner was given, whereas the <em>Cd </em>for a flat plate is generally found to be around <em><u>1.28</u></em><em> </em>which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
  4. Power requirement <u>P</u> for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

  1. v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
  2. The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
  3. The power can be determined this equation: P = F . v, where F represents <em>the drag-force</em> that we will need to determine and v represents the<em> velocity of the airplane</em>
  4. The equation to determine drag-force is: F = 1/2 * d *  C_d * A

In the drag-force equation Cd represents the c<em>o-efficient of drag</em> and A represents the <em>frontal area of the banner/plate/balloon (the object being towed)</em>

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

<u>Part a)</u> We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = <em>1/2 * 0.06* 1.225 * 20</em> = 0.735 N. Now to find the power P we will use P = F . v i.e.<em> 0.735 * 41.66</em> = <u><em>30.62 W</em></u>

<em></em>

<u>Part b) </u>For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = <em>1/2 * 1.28 * 1.225 * 20</em> = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = <u><em>653.225 W</em></u>

<u><em></em></u>

<u>Part c)</u> The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

<u>Part d)</u> The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : <em>pi* r^2 (r = d / 2).</em> Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = <u><em>40.08 W</em></u>

4 0
3 years ago
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