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7nadin3 [17]
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
Physics
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

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A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight
photoshop1234 [79]

Answer:

t=1.9 sec

Explanation:

From the question we are told that:

Height h=28m

Time t=3s

Generally the Newton's equation for Initial velocity upward is mathematically given by

 s=ut+\frtac{1}{2}at^2

 28=3u-\frac{1}{2}*9.8*3^2

 u=24.03m/s

Generally the velocity at  elevation and depression occurs  as ball arrives and passes through S=28

 v=\sqrt{24.03-2*9.8*28}

 v=5.35m/s and -5.35m/s

Generally the Newton's equation for time to reach initial velocity  is mathematically given by

 v=u+at

 5.35=24.03-9.8t

 t=\frac{28.03-5.35}{9.8}

 t=1.9 sec

4 0
3 years ago
A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is
Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2

The acceleration of the craft should be 1.02234 m/s²

F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N

Weight of the craft

W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N

Thrust

F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

3 0
3 years ago
17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106
sleet_krkn [62]

Answer:

Explanation:

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied /  d

= 5 x 10³  / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than  breakdown strength for air  3.0×10⁶ V/m

b ) Let the plates be at a separation of d .so

5 x 10³ / d = 3.0×10⁶ ( break down voltage )

d = 5 x 10³  / 3.0×10⁶

= 1.67 x 10⁻³ m

= 1.67 mm.

5 0
3 years ago
Why were phonograph records placed on the sides of the two voyager spacecraft launched in the 1970s?
LenaWriter [7]

Answer:

The reason for phonograph installation is to provide information about humans to interstellar aliens.

Explanation:

Phonographs are archaic forms of gramophone which are able to record and reproduce sound.Phonographs were used in early days in voyagers to help transmit information about humans to nearby alien species in space.

7 0
3 years ago
The board sandwiched between two other boards shown below weighs 87.5 n. if the coefficient of friction between the boards is 0.
IRINA_888 [86]
Refer to the diagram shown below.

W = 87.5 N, the weight of the sandwiched board.
μ = 0.622, the static coefficient of friction.

From the free body diagram of the sandwiched board, obtain
2μF = W
F = W/(2μ) = 87.5/(2*0.622) = 70.34 N

Answer: 70.34 N

8 0
3 years ago
Read 2 more answers
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