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3 years ago

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An automatic clothing drier spins at 51.6 rev/min. If the radius of the drier drum is 30.5cm, how fast (in m/s) is the drier dru

m moving

1 answer:

Jlenok [28]3 years ago

3 0

Answer:

1.648 m/s

Explanation:

1 revolution equals 2pi radians.

Calculate the angular velocity by taking 2pi x v, then divide by 60 seconds.

To convert this to m/s, simply take this answer and multiply it by 0.305m (a.k.a. the radius of the circle).

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Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram

Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

Force 1 = 500 Newton

Distance 1 = 1.2 meter

Force 2 = 500 Newton

Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

$Work\;done = Force \times distance$

<u>For </u><u>student 1</u><u>:</u>

$Work\;done = 500 \times 1.2$

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

$Work\;done = 500 \times 5$

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

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3 years ago

Read 2 more answers

Choose either eclipses or lunar phases and make a model of the event. Your model can take any form, providing it can explain how

babymother [125]

Answer: See the explanation below.

Explanation: For this assignment, I chose to display how eclipses are created.

My model was made utilizing a 3D displaying device program for all intents and purposes. The items utilized are three models I made for this presentation, Earth, the moon, and the sun. These three models will be utilized for the showcase.

The light that shines from the sun would create a shadow on the moon. The moon would then catch the light that should've arrived on Earth, making the shadow we call an eclipse. Earth gets a shadow of the moon and the remainder of Earth is lit up from the rest of the light, making an eclipse.

The individual I demonstrated my project to was [<em>Someone you know</em>], [<em>Pronoun</em>] said it precisely took after the occasion of an eclipse. The light from the sun being shined on to the moon rather than the Earth, creating the shadow we call an eclipse.

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3 years ago

Any unit of acceleration MUST have the form of _____.

KengaRu [80]

Any unit of acceleration must have the dimensions (form) of

(a unit of length) / (a unit of time)²

7 0

3 years ago

The force an ideal spring exerts on an object is given by Fx = -kx, where x measures the displacement of the object from its equ

shusha [124]

Answer:

The work done by this force can be found via the following formula

$W = \int{F(x)} \, dx = \int\limits^0_{-20} {(-kx)} \, dx = \frac{-kx^2}{2}\left \{ {{x=0} \atop {x=-20}} \right. = \frac{-60*(-20)^2}{2} \\W = -12000J$

Explanation:

Alternatively, the work done by the object is equal to the elastic potantial energy done by the spring.

$U = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 =0 - \frac{1}{2}60(-20)^2 = -12000J$

6 0

3 years ago

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

a

$\theta _2 = 13^o$

b

$\theta _1 =32.94^o$

c

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

So

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

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3 years ago