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ZanzabumX [31]
2 years ago
12

A wheel turns through 5.5 revolutions while being accelerated from rest at 20rpm/s.(a) What is the final angular speed ? (b) How

long does it take to turn the 5.5 revolutions?
Physics
1 answer:
alina1380 [7]2 years ago
3 0

Answer:

(a) The final angular speed is 12.05 rad/s

(b) The time taken to turn 5.5 revolutions is 5.74 s

Explanation:

Given;

number of revolutions, θ = 5.5 revolutions

acceleration of the wheel, α = 20 rpm/s

number of revolutions in radian is given as;

θ = 5.5 x 2π = 34.562 rad

angular acceleration in rad/s² is given as;

\alpha = \frac{20 \ rev}{min} *\frac{1}{s} *(\frac{2\pi \ rad}{1 \ rev } *\frac{1 \ min}{60 \ s}) \\\\\alpha = 2.1 \ rad/s^2

(a)

The final angular speed is given as;

\omega _f^2 = \omega_i ^2 + 2\alpha \theta\\\\\omega _f^2 = 0 +  2\alpha \theta\\\\\omega _f^2 =   2\alpha \theta\\\\\omega _f = \sqrt{2\alpha \theta}\\\\ \omega _f  = \sqrt{2(2.1) (34.562)}\\\\ \omega _f = 12.05 \ rad/s

(b) the time taken to turn 5.5 revolutions is given as

\omega _f = \omega _i + \alpha t\\\\12.05 = 0 + 2.1t\\\\t = \frac{12.05}{2.1} \\\\t = 5.74 \ s

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Which of the following cars have the most kinetic energy
faltersainse [42]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2>

\huge\boxed{OptionA}

<h2>_____________________________________</h2><h2>DATA:</h2><h3>Blue Car: </h3>

mass = 4 kg

velocity = 5 m/s^2

<h3 /><h3>Orange truck:</h3>

Mass= 2kg

Velocity = 7m/s^2

<h3 /><h3>Grey Car:</h3>

mass = 6 kg

velocity = 4m/s^2

<h3 /><h3>Green Car:</h3>

Mass = 8 kg

Velocity = 3 m/s^2

<h2>_____________________________________</h2><h2>SOLUTION:</h2>

By the equation of kinetic energy,

                                       

                                         K.E = \frac{1}{2} mv^2

Where,

            K.E is kinetic energy

            m is mass

            v is velocity

<h2>_____________________________________</h2><h3>Kinetic energy of Blue car:</h3>

 

Directly substitute the variables in the equation,

                                       

                                       K.E = \frac{1}{2}x4x5^2

Simplify the equation,

                                       K.E = 50 J

<h2>_____________________________________</h2><h3>Kinetic Energy of Silver Car:</h3>

                                           

 Directly substitude the variable in the equation,

                                                           

                                        K.E = \frac{1}{2}x6x4^2

Simplify the equation,

                                        K.E = 48J

<h2>_____________________________________</h2><h3>Kinetic Energy of Green Car:</h3><h3 />

Substitute the variables in the equation,

                                         

                                         K.E = \frac{1}{2}x8x3^2

Simplify the Equation,

                                         

                                         K.E = 36J

<h2>_____________________________________</h2><h3>Kinetic Energy of Orange Truck:</h3><h3 />

Substitute the variable,

                                        K.E = \frac{1}{2}x 2x7^2

Simplify the equation,

                                     

                                        K.E = 49J

<h2>_____________________________________</h2>

As you can see that the highest value of kinetic energy is of Blue SUV thus it will be out answer.

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h3 /><h3 /><h3 /><h3 /><h2 /><h2 />
6 0
2 years ago
Only 35 % of the intensity of a polarized light wave passes through a polarizing filter. What is the angle between the electric
Nana76 [90]

Answer:

The angle between the electric field and the axis of the filter is 54⁰

Explanation:

Apply the equation for intensity of light through a polarizer.

I = I_oCos^2 \theta

where;

I is the intensity of the transmitted light

I₀ is the intensity of the incident light

θ is the incident angle

If only 35 % of the intensity of a polarized light wave passes through a polarizing filter, then the ratio of the intensity of the transmitted light to that of the intensity of the incident light is given by;

\frac{I}{I_o}  = Cos^2 \theta\\\\\frac{35}{100} =  Cos^2 \theta\\\\Cos^2 \theta = 0.35\\\\Cos\theta = \sqrt{0.35} \\\\Cos\theta = 0.5916\\\\\theta = Cos^{-1}(0.5916)\\\\\theta  = 54 ^0

Therefore, the angle between the electric field and the axis of the filter is 54⁰

3 0
2 years ago
The change in motion (acceleration) of an object depends on
7nadin3 [17]

Answer:

BOTH the size of the force AND the mass of the object

Explanation:

Acceleration of an object is the rate of change of its velocity.

The relation between force, mass and acceleration is given by the formula as follows :

F = ma

m is mass

a is acceleration

It would mean that the change in motion or the acceleration of an object depends on both the size of the force and the mass of the object. Hence, the correct option is (c).

8 0
3 years ago
A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o
Nataly_w [17]

Explanation:

The given data is as follows.

      C_{1} = 1.50 \times 10^{-6} F

      C_{1} = 3.50 \times 10^{-6} F    

      Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

    \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}

    \frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}

                 = 0.945 \times 10^{-6} F

          C = 1.06 \times 10^{-6} F

Now, we will calculate the charge across each capacitance as follows.

              Q = CV

                  = 1.06 \times 10^{-6} F \times 2.50 V

                  = 2.65 \times 10^{-6} C

                  = 2.65 \mu C

Thus, we can conclude that 2.65 \mu C is the charge stored on each given capacitor.

5 0
3 years ago
A spherical shell has inner radius Rin and outer radius Rout. The shell contains total charge Q, uniformly distributed. The inte
Tju [1.3M]

Answer:

  E = Q / 4 π ε₀  (r-R_int) / (R_put³ -R_int³)

Explanation:

For this exercise we can use Gauss's law

          Ф = ∫ E. dA = q_{int} / ε₀

Where q is the charge inside the surface.

In this case the surface must be a sphere, the electric field lines and the radius of the sphere are parallel, so the scalar product is reduced to the algebraic product

           ∫ E dA = q_{int}/ ε₀

The area of ​​a sphere is

           A = 4π r²

- The electric field for a distance r < R_int

The charge inside is zero, so the electric field

          E = 0        r <R_in t

- The field for a radio inside the shell

   Let's use the concept of density

         ρ = Q / V

         q = ρ (4/3 π r³)

         dq = ρ 4π r² dr

We substitute in the Gaussian equation

     E ∫ dA = ρ 4π r² dr / ε₀

    E 4π r² = ρ 4π/ε₀   r³ / 3

     E = ρ / 3ε₀ r

We evaluate between the lower limit r = R_int, E = 0 and the upper limit r = r, E = E

      E- 0 = ρ / 3ε₀ (r –R_int)

Density  is

      ρ = q / 4/3 π (R_out³ - R_int³)

Where R <r

     E = Q / 4 π ε₀  (r-R_int) / (R_put³ -R_int³)

5 0
3 years ago
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