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3 years ago

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A Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. T

he clock hand has a length of 15.0 cm and a mass of 6.00 g. Take the second hand to be a slender rod rotating with constant angular velocity about one end. Express your answer in kilogram-meters squared per second.

1 answer:

julia-pushkina [17]3 years ago

6 0

Answer:

L = 1.41 x 10⁻⁵ kg.m²/s

Explanation:

First, we need to find the linear speed of the second hand of the clock. Since, we know that second hand completes one revolution in 60 seconds. Therefore, its angular velocity will be:

ω = (1 rev/60 s)(2π rad/ 1 rev)

ω = 0.105 rad/s

Now, for linear speed we use formula:

v = rω

where,

v = linear speed = ?

r = radius of circle = length of hand = 15 cm = 0.15 m

Therefore,

v = (0.15 m)(0.105 rad/s)

v = 0.016 m/s

Now, to find the angular momentum of second hand, we use the following formula:

L = m v r

where,

L = Angular Momentum of Second-Hand = ?

m = mass of second-hand = 6 g = 0.006 kg

Therefore,

L = (0.006 kg)(0.016 m/s)(0.15 m)

<u>L = 1.41 x 10⁻⁵ kg.m²/s</u>

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During their physics field trip to the amusement park, Tyler and Maria took a rider on the Whirligig. The Whirligig ride consist

blondinia [14]

Answer:

Speed = 7.04434 m/s

Explanation:

Given that:

The radius of the circle = 6.5 m

The expression for the calculation of circumference is shown below as:

$Circumference=2\times \pi \times radius$

Thus,

$Circumference=2\times \frac{22}{7} \times 6.5\ m=40.8572\ m$

Time taken = 5.8 seconds.

Thus, The speed is:

$Speed=\frac{Circumference}{Time\ Taken}$

So,

$Speed=\frac{40.8572}{5.8}\ m/s$

<u>Speed = 7.04434 m/s</u>

8 0

4 years ago

Read 2 more answers

A 1.0-kilogram object moving east with a velocity of 10 meters per second collides with a 0.50-kilogram object that is at rest.

NeTakaya

Answer:

10kg m/s

Explanation:

We can use the conversation of momentum for this question. Essentially the momentum before and after the crash will remain the same. We can use the formula P = mv to solve.

P = 1 * 10

P = 10kg m/s

Best of Luck!

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3 years ago

Car B has the same mass as Car A but is going twice as fast. If the same constant force brings both cars to a stop, how far will

evablogger [386]

Answer:four times

Explanation:

Given

mass of both cars A and B are same suppose m

but velocity of car B is same as of car A

Suppose velocity of car A is u

Velocity of car B is 2 u

A constant force is applied on both the cars such that they come to rest by travelling certain distance

using to find the distance traveled

where, v=final velocity

u=initial velocity

a=acceleration(offered by force)

s=displacement

final velocity is zero

For car A

$0-(u)^2=2\times a\times s$

$s_a=\frac{u^2}{2a}------1$

For car B

$0-(2u)^2=2\times a\times s_b$

$s_b=\frac{4u^2}{2a}----2$

divide 1 and 2 we get

$\frac{s_a}{s_b}=\frac{1}{4}$

thus $s_b=4\cdot s_a$

distance traveled by car B is four time of car A

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3 years ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

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4 years ago

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A spherical mirror gives an image magnified 5 times at a distance 5 m. determine whether the mirror is convex or concave? How mu

irakobra [83]

Answer:

1. Concave mirror.

2. 4.17 m or 417 cm.

Explanation:

The following data were obtained from the question:

Object distance (u) = 5 m

Magnification (M) = 5

Focal length (f) =..?

1. Identification of the mirror.

To determine whether or not the mirror is concave or convex, we must first of all calculate the image distance.

This can be obtained as follow:

Object distance (u) = 5 m

Magnification (M) = 5

Image distance (v) =.?

Magnification (M) = image distance (v) /object distance (u).

M = v/u

5 = v/5

Cross multiply

v = 5 x 5

v = 25 m

Since the image distance obtained is positive, the mirror is said to be a concave mirror.

2. Determination of the focal length of the mirro.

This can be obtained as follow:

Object distance (u) = 5 m

Image distance (v) = 25 m

Focal length (f) =...?

1/f = 1/v + 1/u

1/f = 1/25 + 1/5

1/f = 0.04 + 0.2

1/f = 0.24

Cross multiply

f x 0.24 = 1

Divide both side by 0.24

f = 1/0.24

f = 4.17 m

Converting the focal length of cm, we have:

1 m = 100 cm

Therefore, 4.17 m = 4.17 x 100 = 417 cm

Therefore, the focal length of the mirror is 4.17 or 417 cm.

4 0

3 years ago