Ask question

Ask question

All categories

1 year ago

6

In the lab downstairs physics majors use a rotating mirror to measure the speed of light within a few percent of the actual valu

e they switch the mirror from rotating clockwise at 115 rad/s to counter clockwise at 115 rad/s over the course of 85 seconds. Assume constant angular acceleration.
QUESTION: how many revolutions does the rotating mirror go through when it gets to a 0 angular velocity along the way to new speed. SHOW WORKING OUT

1 answer:

iris [78.8K]1 year ago

7 0

The number of complete cycles the rotating mirror goes through before the angular velocity gets to zero is approximately 1166.8 revs

<h3>What is angular velocity?</h3>

Angular velocity is the ratio of the angle turned to the time taken.

The kinematic equation for angular velocity are presented as follows;

ω = ω₀ + α·t

θ = θ₀ + ω₀·t + 0.5·α·t²

Where;

θ₀ = The initial angle turned = 0

ω₀ = The initial angular velocity of the mirrors = 115 rad/s clockwise

α = The angular acceleration = (115 - (-115))rad/s/(85 s) = -46/17 m/s²

t = The duration of the motion;

When the angular velocity, ω is zero, we get;

0 = 115 - 46/17·t

t = 85/2

Which indicates;

θ = 0 + 115× (85/2) + 0.5×(46/17) ×(85/2)² = 7331.25

θ = 7331.25 radians

θ = 7331.25/(2×π) ≈ 1166.8 rev

The mirrors would have turned through approximately 1166.8 revolutions when the angular gets to zero

Learn more about angular velocity and acceleration here:

brainly.com/question/13014974

#SPJ1

You might be interested in

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

3 years ago

Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f

musickatia [10]

Answer:

Er = 231.76 V/m, 27.23° to the left of E1

Explanation:

To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:

E1:

$E_1_x = 0V/m\\E_1_y=100V/m$

E2:

Keep in mind that the x component of electric field E2 is directed to the left.

$E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m$

∑x: $E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m$

∑y: $E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m$

The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.

$||E_r||= \sqrt{(-106.07V/m)^2+(206.07V/m)^2} = 231.76 V/m\\\\\alpha = arctan(\frac{206.7 V/m}{-106.07 V/m}) = 117.24degrees$

or 27.23° to the left of E1.

8 0

3 years ago

the power rating of an electric lawn mower is 2200 watts. if the lawn mower was used for 30 mins, and 650 Newtons of force was u

Tasya [4]

Answer:

Calculate the work done by a 47 N force pushing a 0.025 kg pencil 0.25 m ... A boy on a bicycle drags a wagon full of newspapers at 0.80 m/s for 30 min ... A power mower does 9.00 x 105 J of work in 0.500 h. ... p: W 2200ch: w will320,000 T/ ... How much electrical energy (in kilowatt hours) would a 60.0 W light bulb use in ..

Explanation:

4 0

2 years ago

The measure of the force with which air molecules push on a surface is called

alex41 [277]

It is called air pressure

7 0

3 years ago

What property of light shown in the picture?

kupik [55]

Answer:

Explanation:

Reflection

8 0

2 years ago

Read 2 more answers