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3 years ago

A salt shaker sits 0.102 m from

Physics

1 answer:

Ivahew [28]3 years ago

6 0

The maximum speed is 0.55 m/s

Explanation:

For an object in uniform circular motion, the force of friction between the object and the ground provides the centripetal force required to keep the body in motion. Therefore we can write:

$\mu_s mg = m\frac{v^2}{r}$

where the term on the left is the frictional force and the term on the right is the centripetal force, and where

$\mu_s$ is the coefficient of static friction

m is the mass of the body

g is the gravitational acceleration

v is the speed of the body

r is the radius of the circular path

In this problem, we have:

$\mu_s = 0.307$

r = 0.102 m

$g=9.8 m/s^2$

Substituting and re-arranging, we find the maximum speed v at which the salt shaker can rotate:

$v=\sqrt{\mu gr}=\sqrt{(0.307)(9.8)(0.102)}=0.55 m/s$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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Yes, yes, we know all of that. It certainly took you long enough to
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If
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7 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

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3 years ago

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Anna11 [10]

Answer:

4.8 mph

Explanation:

From the question,

Average speed = total distance/total time

V = d/t....................... Equation 1

Where d = distance, t = time

Given: d = 26.2 miles, t = 5.5 hours.

Substitute these values into equation 1

V = 26.2/5.5

V = 4.76 mph

V ≈ 4.8 mph

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3 years ago

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Answer:

5.25%

Explanation:

From the question given above, the following data were obtained:

Accepted value = 238857 miles

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Next, we shall determine the absolute error. This can be obtained as follow:

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Absolute Error =?

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