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2 years ago

What are the4 spheres on earth

Physics

2 answers:

bezimeni [28]2 years ago

6 0

Answer:

Hydrosphere, Lithosphere, Atmosphere and Biosphere

Anon25 [30]2 years ago

5 0

Answer:

northen henisphere,southern hemisphere, Eastern hemisphere, Western hemisphere.

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Which of the following best describes the circuit shown below?

Rudiy27

Answer:

Short circuit

Explanation:

The given figure shows a short circuit. It is defined as the circuit which allows the flow of electric current when there is no resistance. It shows a battery, bulb and connecting wires.

The wire across the bulb is connected from one terminal to another without any resistance in between them.

So, the correct option is (d) " short circuit ".

3 0

2 years ago

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THE PROBLEM!

Elodia [21]

Answer: 15.29

Explanation: there you go have a nice day (*^p^*)

3 0

2 years ago

(FIRST TO ANSWER CORRECTLY WILL BE THE BRAINIEST!!!)How will the temperatures of the water in the beakers compare if an equal am

bulgar [2K]

Answer:

the answer is the temperatures of both beakers' water will increase by the same amount...

Explanation:

I know this because i just did it on study island

7 0

2 years ago

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The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental

pav-90 [236]

A) $5.0\cdot 10^{-11} m$

The energy of an x-ray photon used for single dental x-rays is

$E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J$

The energy of a photon is related to its wavelength by the equation

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

Re-arranging the equation for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m$

B) $2.0\cdot 10^{-11} m$

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

$E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J$

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m$

4 0

2 years ago

Determine the weight of an average physical science textbook whose mass is 3.1 kilograms. The acceleration due to gravity is 9.8

Mrrafil [7]

Given:

mass is 3.1 kilograms

The acceleration due to gravity is 9.8m/s2

Required:

Weight

Solution:

W = mg

W = (3.1 kilograms)( 9.8m/s2)

W = 30.38 Newtons

5 0

2 years ago