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rodikova [14]
3 years ago
8

20 POINTS!! What does this picture look like ???

Physics
1 answer:
oksian1 [2.3K]3 years ago
6 0
A water toy that brings water up and down I’m not sure what it’s called
I think but I’m not sure
You might be interested in
Mass (kg) 4.0
densk [106]

Answer:

25 m/s

Explanation:

First of all, we can find the acceleration the object by using Newton's second law of motion:

F=ma

where

F = 20.0 N is the net force applied on the object

m = 4.0 kg is the mass of the object

a is its acceleration

Solving for a, we find

a=\frac{F}{m}=\frac{20}{4}=5.0 m/s^2

Now we know that the motion of the object is a uniformly accelerated motion, so we can find its final velocity by using the following suvat equation:

v=u+at

where

v is the final velocity

u = 0 is the initial velocity

a=5.0 m/s^2 is the acceleration

t = 5 s is the time

By substituting,

v=0+(5.0)(5)=25 m/s

8 0
3 years ago
An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t
Fofino [41]

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

P(z)=P_{0}+\rho _wgh

Now the gauge pressure is given by

P(z)-P_{0}=\rho _wgh

Applying values we get

P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa

8 0
3 years ago
Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is 2500 W/m2, of which
GuDViN [60]

Answer:

i. 0.34

ii. 0.4

iii. 1700 w/m²

iv. 2211.36 w/m²

Explanation:

Given that

Irradiation of the plate, G = 2500 w/m²

Reflected rays, p = 500 w/m²

Emissive power, E = 1200 w/m²

See attachment for calculations

8 0
4 years ago
When atoms of an element are excited, they emit specific wavelengths of light. How is this similar to a fingerprint when Fraunho
Anika [276]

Answer:

As you may know, each element has a "fixed" number of protons and electrons.

These electrons live in elliptical orbits around the nucleus, called valence levels or energy levels.

We know that as further away are the orbits from the nucleus, the more energy has the electrons in it. (And those energies are fixed)

Now, when an electron jumps from a level to another, there is also a jump in energy, and that jump depends only on the levels, then the jump in energy is fixed.

Particularly, when an electron jumps from a more energetic level to a less energetic one, that change in energy must be compensated in some way, and that way is by radiating a photon whose energy is exactly the same as the energy of the jump.

And the energy of a photon is related to the wavelength of the photon, then we can conclude that for a given element, the possible jumps of energy levels are known, meaning that the possible "jumps in energy" are known, which means that the wavelengths of the radiated photons also are known. Then by looking at the colors of the bands (whose depend on the wavelength of the radiated photons) we can know almost exactly what elements are radiating them.

7 0
3 years ago
"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in gro
pickupchik [31]

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called  finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

5 0
3 years ago
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