Answer:
Explanation:
⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆
mass defect = 56.936296 + .00055 - 56.935399
= .001447 u
equivalent energy
= 931.5 x .001447 MeV
= 1.3479 MeV .
= 1.35 MeV
energy of gamma ray photons = .14 + .017
= .157 MeV .
Rest of the energy goes to neutrino .
energy going to neutrino .
= 1.35 - .157
= 1.193 MeV.
Answer:
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
Explanation:
For this exercise let's use the relationship between momentum and momentum.
I = F t = Δp
in this case the final velocity is zero
F t = 0 -m v₀
F = m v₀ / t
in order to answer the question we must assume that the two vehicles have the same mass and speed
concrete barrier
F₁ = -p₀ / 0.1
F₁ = - 10 p₀
barrier collapses
F₂ = -p₀ / 1
let's look for the relationship of the forces
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 2, 6, and 75, and all planets will orbit the sun successfully.
If the mass of the sun is 2x, at least one planet will fall into the habitable zone. if I place a planet in orbits 84, 1, and 5, and all planets will orbit the sun successfully.
If the mass of the sun is 3x, at least one planet will fall into the habitable zone if I place a planet in orbits 672, and 7 and all planets will orbit the sun successfully.
Explanation:
It is given that,
Mass of the soccer ball, m = 0.425 kg
Speed of the ball, u = 15 m/s
Angle with horizontal, 
Time for which the player's foot is in contact with it, 
Part A,
The x component of the soccer ball's change in momentum is given by :
The y component of the soccer ball's change in momentum is given by :
Hence, this is the required solution.
Answer:
The friction force is 250 N
Explanation:
The desk is moving at constant velocity. This means that its acceleration is zero: a = 0. Newton's second law states that the resultant of the forces acting on the desk is equal to the product between mass (m) and acceleration (a):
In this case, we know that the acceleration is zero: a = 0, so also the resultant of the forces must be zero:
(1)
We are only interested in the forces acting along the horizontal direction, since it is the direction of motion. There are two forces acting in this direction:
- the pull, forward, F = 250 N
- the friction force, backward, 
Given (1), we have
So the force of friction must be equal to the pull:
