Answer:
Please find the answer in the explanation
Explanation:
1.) How far is Object Z from the origin at t = 3 seconds
The distance of the object Z from the origin will be the slope of the graph.
Slope = 4/2 = 2m
2.) Which object takes the least time to reach a position 4 meters from the origin ?
According to the graph given to the question above, object Z has the list time which is 2 seconds since object X does not start from the origin.
3.) Which object is farthest from the origin at t = 2 seconds?
The correct answer is still object Z because it has the highest slope.
We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is
From the question we are told
Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e
<h3>An E
xpression for the <em>magnitude </em>of charge moved</h3>
Generally the equation for the <em>magnitude </em>of charge moved, Q is mathematically given as
Therefore
An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be
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An impact which stops a moving object must do enough work to take away its kinetic energy, so extending the distance moved during the collision reduces the impact force.
Atmosphere
Atmospheric gas from prehistoric eras is found trapped in glaciers in the form of bubbles. These gas bubbles are the basis of studying ice cores as they provide us with accurate estimates of the conditions of past climates. The bubbles allow us to determine the composition of atmospheric air, such as the carbon dioxide and methane concentrations, as well as allow us to determine air temperatures in the past.
This question apparently wants you to get comfortable
with E = m c² . But I must say, this question is a lame
way to do it.
c = 3 x 10⁸ m/s
E = m c²
1.03 x 10⁻¹³ joule = (m) (3 x 10⁸ m/s)²
Divide each side by (3 x 10⁸ m/s)²:
Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)
= (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)
= 1.144 x 10⁻³⁰ kg . (choice-1)
This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature. The question is just an
exercise in arithmetic, and not a particularly interesting one.
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Something like this could have been much more impressive:
The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?
Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)
= (2,242 x 10⁶ x 86,400 x 365) joules
= 7.0704 x 10¹⁶ joules .
How much converted mass is that ?
E = m c²
Divide each side by c² : Mass = E / c² .
c = 3 x 10⁸ m/s
Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)
= 0.786 kilogram ! ! !
THAT should impress us ! If I've done the arithmetic correctly,
then roughly (1 pound 11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !
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