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3 years ago

The frequency of a generated electromagnetic wave matches the

Physics

1 answer:

Assoli18 [71]3 years ago

8 0

Answer:

C) frequency of vibrating charges that produced it.

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8. Where are the ribosomes usually located in plant and animal cells?

professor190 [17]

Answer:

Ribosomes are found 'free' in the cytoplasm or bound to the endoplasmic reticulum (ER) to form rough ER.

5 0

3 years ago

Show that 1.0 m/s = 3.6 km/h . Hint: Show the explicit steps involved in converting 1.0 m/s = 3.6 km/h.

sergiy2304 [10]

Just multiply the "1.0 m/s" by ' 1 ' a few times. (Remember that a fraction with the same quantity on top and bottom is equal to ' 1 ' .)

(1.0 m/sec) · (1 km/1000 m) · (60 sec/min) · (60 min/hr) =

(1.0 · 60 · 60 / 1,000) (m · km · sec · min / sec · m · min · hr) =

(3,600 / 1,000) (km / hr) =

3.6 km/hr .

5 0

4 years ago

A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular

ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

$\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0$

Where,

$\theta =$ Angular Displacement

$\alpha =$ Angular Acceleration

$\omega_0 =$ Angular velocity

$\theta_0 =$ Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

$\theta = \frac{1}{2}\alpha t^2$

Re-arrange for $\alpha,$

$\alpha = \frac{2\theta}{t^2}$

Replacing our values,

$\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}$

$\alpha = 0.139rad/s$

Therefore the ANgular acceleration of the mass is $0.139rad/s^2$

4 0

3 years ago

How do your results from ray tracing compare to your results from using the thin-lens equation?

EastWind [94]

Answer:

20cm

Explanation:

A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).

According to the lens equation

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ where;

f is the focal length of the lens

u is the object distance

v is the image distance

If the magnification is - 0.6

mag = v/u = -0.5

v = -0.5u

since v = 10cm

10 = -0.5u

u = -10/0.5

u =-20 cm

Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{1}{-20} + \frac{1}{10}\\\frac{1}{f} = \frac{-1+2}{20} \\\frac{1}{f} = \frac{1}{20} \\cross \ multiply\\f = 20\\f = 20 cm$