210J
PE is mgh in this context.
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>
Answer:
Significant Other
Explanation:
that comes to mind when I see that
Answer:
Explanation:
5 C = 278 K
25 C = 298 K
V1 / T1 = V2 / T2
1.5L / 278 K = V2 / 298 K
V2 = (1.5L * 298) / 278
V2 = 1.61 L
Answer:
size of popcorn seed may meet the scaled earth size of x = 0.07312 in.
Explanation:
Given:-
- The diameter of actual sun, Ds = 1.3927 * 10^6 km
- The diameter of model sun ball, ds = 8 in
- The diameter of the actual earth, De = 12,742 km
Find:-
How large would be the earth model
Solution:-
- We can use the actual diameters of the earth and sun to develop the ratio of earth diameter to that of sun. The direct ratio can be used to model the diameter of earth for the scale model
Sun Earth
Actual : 1392700 km 12,742 km
Scale : 8 in x
- Use direct relations and solve for x:
x = (8 in)*(12,742 km ) / ( 1392700 km )
x = 0.07312 in.
- From the given options apple, grapefruit and marble diameters are larger than 1 inch. So only the size of popcorn seed may meet the scaled earth size of x = 0.07312 in.
