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Veseljchak [2.6K]
4 years ago
10

The 1.00-cm-long second hand on a watch rotates smoothly.

Physics
1 answer:
swat324 years ago
6 0

Answer:

Angular velocity = 0.105 rads⁻¹

speed of tip = 0.00105 ms⁻¹

Explanation:

A second hand turns  2π rad each 60 sec

Thus: it has an angular speed given by

ω = 2π/T = 2π/60 = π/30 = 0.105 rad s⁻¹

The speed of the tip is given by

V= rω = 0.01 m × (0.105 rads⁻¹)

= 0.00105 ms⁻¹

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A student runs up a flight of stairs which info is not needed to calculate the rate of the student is doing work against gravity
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Answer:

B.

Explanation:

Given that a student runs up a flight of stairs which info is not needed to calculate the rate of the student is doing work against gravity A the height of the flight of stairs B the length of flight of stairs C the time taken to run up the stairs D the weight of the student

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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 14 s until its motor stops. Disreg
telo118 [61]

Answer:

The maximum height of the rocket will be 1.0 × 10⁴ m.

Explanation:

Hi there!

The height of the rocket at time "t" can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · a · t²  (when the rocket has an upward acceleration)

y = y0 + v0 · t + 1/2 · g · t²  (after the motor of the rocket stops)

Where:

y = height.

y0 = initial height.

v0 = initial velocity.

t = time.

a = acceleration due to the motors.

g = acceleration due to gravity.

The velocity of the rocket can be calculated as follows:

v = v0 + a · t  (while the motor is running)

v = v0 + g · t  (after the motor stops)

Where "v" is the velocity of the rocket at time "t".

The rocket rises with upward acceleration for 14 s. After that, the rocket starts being accelerated in the downward direction due to gravity. But it will continue going up after the motor stops because the rocket has initially an upward velocity that will be reduced until it becomes zero and the rocket starts to fall.

Let´s find the height reached by the rocket while it was accelerated in the upward direction (the origin of the frame of reference is located at the launching point and upward is the positive direction):

y = y0 + v0 · t + 1/2 · a · t²

y = 0 m + 0 m/s · 14 s + 1/2 · 28 m/s² · (14 s)²

y = 2.7 × 10³ m

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v = v0 + g · t

0 = 3.9 × 10² m/s - 9.8 m/s² · t

-3.9 × 10² m/s/ -9.8 m/s² = t

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After the motor stops, it takes the rocket 40 s s to reach the maximum height.

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y = y0 + v0 · t + 1/2 · g · t²

y = 2.7 × 10³ m +  3.9 × 10² m/s · 40 s - 1/2 · 9.8 m/s² · (40 s)²

y = 1.0 × 10⁴ m

The maximum height of the rocket will be 1.0 × 10⁴ m

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