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3 years ago

Approximately how many degrees does high latitude ocean water change from the ocean surface to a depth of 1 km

Physics

1 answer:

saveliy_v [14]3 years ago

3 0

There is no change in ocean water temperature when it is from the ocean surface to a depth of 1 km.

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Calculate the wavelength of the electromagentic waves with the given frequencies, and determine the type of electromagnetic radi

ololo11 [35]

To develop this problem we require the concepts related to wavelength and its expression to calculate it.

The wavelength is given by

$\lambda =\frac{c}{f}$

Where,

$c=3*10^8 m/s$ light velocity

f = frequency.

Our values are given by,

$f_1=4.38*10^{14}Hz$

$f_2= 4.14*10^{20}Hz$

$f_3 = 3.24*10^{12}Hz$

Then,

$\lambda_1=\frac{3*10^8}{4.38*10^{14}}= 6.8493*10^{-7}m$ Visible

$\lambda_2=\frac{3*10^8}{4.14*10^{20}}= 7.24*10^{-13}m$ Gamma Ray

$\lambda_3=\frac{3*10^8}{3.24*10^{12}}= 9.259*10^{-5}m$ Infrared

*Note the designation on the type of rays that are, can be found in consulted via On-line or in the optical books referring to the electromagnetic spectrum table with their respective ranges.

5 0

3 years ago

To control whether an object is solid or incorporeal (things can pass through it) you would use the:

Zarrin [17]

Answer:

Gamma radiation or Cathode rays

Explanation:

by striking incident gamma or cathode rays onto the solid when placed on a photographic plate

5 0

2 years ago

Which group in the periodic table is known as salt farmers?

adell [148]

Which group in the periodic table is known as salt formers?

The correct option is the last one: Halogen family.

You can find the halogen or "salt formers" in the group 17 of the periodic table. These are:

- Fluorine.
-Chlorine.
- Bromine.
- Iodine.
- Astatine.

All of them are non-metallic elements and they have 7 electrons.

5 0

3 years ago

Read 2 more answers

GIVING BRAINLIEST PLEASE HELP!!

OLga [1]

A. because I had this question yesterday.

3 0

3 years ago

A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.10 104 rad/s to an angular speed of 3.14

Anestetic [448]

Answer:

3.63 s

Explanation:

We can solve the problem by using the equivalent SUVAT equations for the angular motion.

To find the angular acceleration, we can use the following equation:

$\omega_f^2 - \omega_i ^2 =2 \alpha \theta$

where

$\omega_f = 3.14\cdot 10^4 rad/s$ is the final angular speed

$\omega_i = 1.10 \cdot 10^4 rad/s$ is the initial angular speed

$\theta= 2.00 \cdot 10^4 rad$ is the angular distance covered

$\alpha$ is the angular acceleration

Re-arranging the formula, we can find $\alpha$ :

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}=\frac{(3.14\cdot 10^4 rad/s)^2-(1.10\cdot 10^4 rad/s)^2}{2(2.00\cdot 10^4 rad)}=2.16\cdot 10^4 rad/s^2$

Now we want to know the time the bit takes starting from rest to reach a speed of $\omega_f=7.85\cdot 10^4 rad/s$ . So, we can use the following equation: