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tamaranim1 [39]

2 years ago

5

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.60 cm from the axis to equal

300000 g (where g is the acceleration due to gravity).
Express your answer numerically in revolutions per minute.

1 answer:

iVinArrow [24]2 years ago

8 0

Answer:

Explanation:

300000(9.8) = ω²(0.0160)

ω = 13555 rad/s

13555 rad/s (60 s/min/ 2π rad/rev) = 129,445 rev/min

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Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

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

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

b)

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0

3 years ago

Qué representa el lanzamiento vertical ascendente

rusak2 [61]

Answer:

Uhhhhh? can u explains? lol

6 0

3 years ago

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

Lostsunrise [7]

Answer:

$0.06\Omega/m$

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire ( $I=\frac{V}{R}$ ). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

$\frac{13}{20z+r} =7.6$ .......(1)

$\frac{13}{40z+r} =4.5$ ......(2)

Substituting the value of r from (2) in (1), we have,

$13=152z+7.6\times\frac{13-180z}{4.5}$

which simplifying gives us, $z=0.0589\Omega/m\approx0.06\Omega/m$ (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 $\Omega$

3 0

3 years ago

A wheel rotates clockwise 6 times per second. What will be its angular displacement after 7 seconds? Answer should be rounded to

solong [7]

Answer:

The frequency of the wheel is the number of revolutions per second:

f= \frac{N_{rev}}{t}= \frac{10}{1 s}=10 Hz

And now we can calculate the angular speed, which is given by:

\omega = 2 \pi f=2 \pi (10 Hz)=62.8 rad/s in the clockwise direction.

Explanation:

5 0

3 years ago

Read 2 more answers

Need help please and thank you

Scilla [17]

C
B
A
C
C

Hope this helps you!!!

8 0

3 years ago