Answer:
You increase the acceleration of the car.
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 !
</span>
Calculate the change in heat of the aluminum; show all calculations. Calculate the change in heat of the water; show all calculations. Are the two values the same? Why or why not? See the attached picture for the numbers.
I got -3443.14 J for the aluminum and 3443.595 for the water
<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg.
The key here is the conservation of momentum.
For the first truck, the momentum is
0(5100 + 4300)
The second truck has a starting momentum of
60(5100 + x)
And finally, after the collision, the momentum of the whole system is
42(5100 + 4300 + 5100 + x)
So let's set the equations for before and after the collision equal to each other.
0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x)
And solve for x, first by adding the constant terms
0(5100 + 4300) + 60(5100 + x) = 42(14500 + x)
Getting rid of the zero term
60(5100 + x) = 42(14500 + x)
Distribute the 60 and the 42.
60*5100 + 60x = 42*14500 + 42x
306000 + 60x = 609000 + 42x
Subtract 42x from both sides
306000 + 18x = 609000
Subtract 306000 from both sides
18x = 303000
And divide both sides by 18
x = 16833.33
So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums.
60(5100 + 16833.33) = 60(21933.33) = 1316000
42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000
They match. The 2nd truck was definitely over loaded.</span>
Option A, current (thumb) to magnetic field (fingers)
As per the First right-hand rule,
Using right hand, if we suppose that thumb points towards the electric current
fingers curl towards the magnetic field
