Answer: the correct answer is 7.8026035971 x 10^(-13) joule
Explanation:
Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get
m = 4.00260 u + 222.01757 u = 226.02017 u .
The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is
Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,
which is equivalent to an energy change of
Delta E = (0.00523 u)*(931.5MeV/1u)
Delta E = 4.87 MeV
Converting 4.87 MeV to Joules
1 joule [J] = 6241506363094 mega-electrón voltio [MeV]
4 mega-electrón voltio = 6.40870932 x 10^(-13) joule
4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Answer:
remains the same, but the apparent brightness is decreased by a factor of four.
Explanation:
A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.
It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).
The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).
The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.
The apparent brightness of a star is measured in watts per square meter.
If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).
Some of the examples of stars are;
- Canopus.
- Sun (closest to the Earth)
- Betelgeuse.
- Antares.
- Vega.
Answer:
The skater's speed after she stops pushing on the wall is 1.745 m/s.
Explanation:
Given that,
The average force exerted on the wall by an ice skater, F = 120 N
Time, t = 0.8 seconds
Mass of the skater, m = 55 kg
It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

The change in momentum is equal to the impulse delivered. So,

So, the skater's speed after she stops pushing on the wall is 1.745 m/s.