PLEASE HELP!!! WILL MARK BRAINLIEST!!!!!!!!!!!!!!!!!!!!

A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a rad

Anna71 [15]

Answer:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

$a_{r} = \omega^{2}\cdot R$

Where:

$\omega$ - Angular speed, measured in radians per second.

$R$ - Radius of rotation, measured in meters.

The angular speed is first determined:

$\omega = \frac{\pi}{30}\cdot \dot n$

Where $\dot n$ is the angular speed, measured in revolutions per minute.

If $\dot n = 3000\,rpm$ , the angular speed measured in radians per second is:

$\omega = \frac{\pi}{30}\cdot (3000\,rpm)$

$\omega \approx 314.159\,\frac{rad}{s}$

Now, if $\omega = 314.159\,\frac{rad}{s}$ and $R = 0.1\,m$ , the resultant acceleration is then:

$a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)$

$a_{r} = 9869.588\,\frac{m}{s^{2}}$

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

$a_{r} = 1006.382g \,\frac{m}{s^{2}}$

6 0

3 years ago

In what way does the sun's rays of light hit the earth?<br><br> PLease anSWER

Dafna11 [192]

Answer:

The Sun's energy gets to the Earth through radiation, which you can prove just by standing outside and letting the sun's rays warm your face on a sunny day. Every object around you is continually radiating unless its temperature is at absolute zero, at which point its molecules completely stop moving.

Hope that helps :)

3 0

3 years ago

Read 2 more answers

Fill in the blank: An object is most likely to sink in water if _________________.

drek231 [11]

High density
random words to fill up 20 character minimum for answering questions :P

4 0

3 years ago

Read 2 more answers

A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them

Tatiana [17]

Answer:

A) q₂ = 75.98 cm, B) q₂' = 115.38 cm, C)

Explanation:

A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.

Lens 1. More to the left

let's use the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively,

We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.

$\frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}$

$\frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}$

1 / q₁ = 0.04756

q₁ = 21.0227 cm

this image is the object for the second lens that has f₂ = 14.8 cm

the distance must be measured from the second lens

p₂ = 39.4 -q₁

p₂ = 39.4 -21.0227

p₂ = 18.38 cm

let's use the constructor equation

1 / q₂ = 1 / f - 1 / p2

$\frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}$

$\frac{1}{q_2}$ = 0.01316

q₂ = 75.98 cm

measured from the second lens

B) the position of the final image with respect to the first lens is

q₂’= q₂ + 39.4

q₂'= 75.98 +39.4

q₂' = 115.38 cm

C) the magnification of a lens is

m = - q / p

in this case the image measured from lens 2 is q2 = 75.98 cm

the distance to the object from the first lens is p1 = 50cm

m = - 75.98 / 50

m = -1.5 X

the negative sign indicates that the image is inverted

4 0

3 years ago

A proton with an initial speed of 400,000 m/s is brought to rest by an electric field. Did the proton move into a region of high

Anton [14]

A proton is initially moving at high speed

$v = 400000 m/s$

so here initially proton is having high speed and high kinetic energy

now this proton comes to rest by external electric field

so the potential energy of proton will increase or we can say that kinetic energy will convert into electrostatic potential energy

so here the proton must have to move at higher potential region

so here correct answer must be

b. Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential.

8 0

3 years ago