Answer: 938
Explanation:
1.67x10^-27kg (938V/c2)
why is gas matter even though we cannot see it?
Answer:
There is matter in gas but due to the high intermolecular forces, all the molecules of gas are far apart
Since the size of a molecule is almost negligible to the human eye, we cannot see the gas molecules and hence, gas is invisible to the naked eye
If you compress a gas against the intermolecular forces, the gas will turn into a liquid
Why is a control group generally very important in an experiment?
Answer: The control group is the part where you see what happens when you change a variable you want to study/examine. Basically, you need the control group because you need something to see what happens when you change something.
Hope this Helps! :))))
In 2006, NASA’s Mars Odyssey orbiter detected violent gas eruptions on Mars, where the acceleration due to gravity is 3.7 m/s2. The jets throw sand and dust about 62.0 m above the surface. Scientists estimate that the jets originate as high-pressure gas speeds through vents just underground at about 130 km/h. How much energy per kilogram of material is lost due to nonconservative forces as the high-speed matter forces its way to the surface and into the air? (Express your answer to two significant figures.)
Answer:
The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.
Explanation:
We can estimate the unit energy losses of gas eruption by Principle of Energy Conservation and Work-Energy Theorem:
[tex]U_{g,1} + K_{1} = U_{g,2}+K_{2}+W_{loss}[/tex] (Eq. 1)
Where:
[tex]U_{g,1}[/tex] - Gravitational potential energy of gas eruptions at surface, measured in joules.
[tex]U_{g,2}[/tex] - Gravitational potential energy of gas eruptions at highest height, measured in joules.
[tex]K_{1}[/tex] - Translational kinetic energy of gas eruptions at surface, measured in joules.
[tex]K_{2}[/tex] - Translational kinetic energy of gas eruptions at highest height, measured in joules.
[tex]W_{loss}[/tex] - Energy losses due to nonconservative forces, measured in joules.
We clear the component associated with energy losses in (Eq. 1):
[tex]W_{loss} = U_{g,1}-U_{g,2}+ K_{1}-K_{2}[/tex]
And we expand it afterwards:
[tex]W_{loss} = m\cdot g\cdot (z_{1}-z_{2}) + \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex] (Eq. 2a)
[tex]w_{loss} = g\cdot (z_{1}-z_{2})+\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex] (Eq. 2b)
Where:
[tex]W_{loss}[/tex] - Energy losses due to nonconservative forces, measured in joules.
[tex]w_{loss}[/tex] - Unit energy losses due to nonconservative forces, measured in joules per kilogram.
[tex]g[/tex] - Gravitational acceleration, measured in meters per second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Bottom and top height, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Gas eruption speeds at surface and highest heights, measured in meters per second.
If we know that [tex]g = 3.7\,\frac{m}{s^{2}}[/tex], [tex]z_{1} = 0\,m[/tex]. [tex]z_{2} =62\,m[/tex]. [tex]v_{1} = 36.111\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the unit energy losses are:
[tex]w_{loss} = \left(3.7\,\frac{m}{s^{2}} \right)\cdot (62\,m-0\,m)+\frac{1}{2} \cdot \left[\left(36.11\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]w_{loss} = 881.40\,\frac{J}{kg}[/tex]
The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.
which of the following units is part of the international system of units, or SI?
Pounds meters
ounces inches
it is meters and ounces
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA's idea is basicallyan electric slingshot that consists of 4 electrodes arranged in a horizontal square with sides of length d at a height h above the ground. The satellite is then placed on the ground aligned with the center of the square. A power supply will provide each of the four electrodes with a charge of Q/4 and the satellite with a charge -Q. When the satellite is released from rest, it moves up and passes through the center of the square. At the instant it reaches the square's center, the power supply is turned offand the electrodes are grounded, giving them a zero electric charge. To test this idea, you decide to use energy considerations to calculate how big Q will have to be to get a 100 kg satellite to a sufficient orbit height. Assume that the satellite startsfrom 15 meters below the square of electrodes and that the sides of the square are each 5 meters. In your physics text you find the mass of the Earth to be 6.0 x 1024kg.
Answer:
The answer is "[tex]q=0.0945\,C[/tex]".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).
[tex]U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}[/tex]
It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:
[tex]U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}[/tex]
Potential energy shifts:
[tex]= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\[/tex]
[tex]=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J[/tex]
Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.
[tex]=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\[/tex]
[tex]=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\[/tex]
[tex]\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C[/tex]
This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
Help pls it’s urgent
Answer:
decreases, but frequency increases.
Explanation:
Wavelength and frequency are inversely proportional, meaning the higher the frequency, the shorter the wavelength, and the lower the frequency, the longer the wavelength.
Which refers to the continuous release of nuclear energy caused when one fission reaction triggers more nuclear reactions?
-fusion reaction
-trigger effect
-energy effect
-chain reaction
Answer:d
Explanation:
Answer:
D is correct
hope it helps
(HURRY I NEED HELP NOW ILL GIVE YOU BRILLIANT). Why do you think it's impossible for an element to appear on only one side of a valid
chemical equation? And I science
Answer:
i got you
Explanation:
Unfortunately, it is also an incomplete chemical equation. ... But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants but only one oxygen atom in the products.
Marta , who is only 5years old , heard her mother use a curse word and is now repeating that word much to the embarrassment of her parents. It was especially upsetting when she used it ik n church! By explaining to Marta that word is not nice and then ignoring her use of the word , her parents are pleased when the word disappeared from her vocabulary. Which concept of learning does this mostly demonstrate?
Protons have ___charge; they have equal amounts of positive and negative ____charges?
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.54 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2230 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.25 V/m, (b) in the negative z direction and has a magnitude of 5.25 V/m, and (c) in the positive x direction and has a magnitude of 5.25 V/m
Answer:
(a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]
(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]
(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]
Explanation:
Given that,
Magnetic field [tex]B=-3.54\times10^{-3}i\ T[/tex]
Velocity = 2230j m/s
We know that,
The net force acting on the proton is equal to the sum of electric and magnetic force.
[tex]F=F_{e}+F_{B}[/tex]
(a). If the electric field is in the positive z direction and has a magnitude of 5.25 V/m,
We need to calculate the magnitude of the net force acting on the proton
Using formula of net force
[tex]F_{net}=e(E+v\times B)[/tex]
Put the value into the formula
[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]
[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]
[tex]F_{net}=(2.1\times10^{-18}\ N)k[/tex]
(b). If the electric field is in the negative z direction and has a magnitude of 5.25 V/m,
We need to calculate the magnitude of the net force acting on the proton
Using formula of net force
[tex]F_{net}=e(E+v\times B)[/tex]
Put the value into the formula
[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]
[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]
[tex]F_{net}=(4.23\times10^{-19}\ N)k[/tex]
(c). If the electric field is in the positive x direction and has a magnitude of 5.25 V/m
We need to calculate the magnitude of the net force acting on the proton
Using formula of net force
[tex]F_{net}=e(E+v\times B)[/tex]
Put the value into the formula
[tex]F_{net}=1.6\times10^{-19}(5.25i+2230\times-3.54\times10^{-3}(j\times i))[/tex]
[tex]F_{net}=(8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]
Hence, (a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]
(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]
(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]
A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find car initial kinetic energy. the final kinetic energy
Answer:
Ek1 = 900000 [J]
Ek1 = 400000 [J]
Explanation:
In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:
[tex]E_{k1}=\frac{1}{2}*m*v1^{2}[/tex]
where:
m = mass = 500 [kg]
v1 = 60 [m/s]
So we have:
Ek1 = 0.5*500*(60^2)
Ek1 = 900000 [J]
and:
Ek2 = 0.5*500*(40^2)
Ek2 = 400000 [J]
A monarchy is the type of government that the colonist do not want true or false
Answer:
Monarchy is rule from kings and queens
Explanation:
Calculate the acceleration of the object in free fall (picture shown above)
Answer:
Estimate slope= acceleration of -18.75 m/s^2
Explanation:
In the velocity vs time graph, acceleration is the slope
Let's take two points from the line and find the slope
(0,0) (0.8, -15)
slope = -15-0/ 0.8-0 = -18.75 m/s^2