What is the pigment in the leaves called?
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<h2><u>KINETIC ENERGY</u></h2><h3>Problem:</h3>
» A 2kg mass is moving at 3m/s. What is its kinetic energy?
<h3>Answer:</h3>
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<h3>Formula:</h3>To calculate the velocity of a kinetic energy, we can use formula
where,
- v is the velocity in m/s
- KE is the kinetic energy in J (joules)
- m is the mass in kg
— — —
Based on the problem, the givens are:
- KE (Kinetic energy) = ? (unknown)
- m (mass) = 2 kg
- v (velocity) = 3 m/s
To get the velocity, substitute the givens in the formula above then solve.
Therefore, the kinetic energy is 9 Joules.
Let us consider two bodies having masses m and m' respectively.
Let they are separated by a distance of r from each other.
As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - where G is the gravitational force constant.
From the above we see that F ∝ mm' and
Let the orbital radius of planet A is = r and mass of planet is
.
Let the mass of central star is m .
Hence the gravitational force for planet A is
For planet B the orbital radius and mass
Hence the gravitational force
Hence the ratio is
[ ans]
Answer:
a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing
b) E_total = 1.89 10⁶ N / C, field is incoming radial
c) E_total = 0
Explanetion:
For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is
Ф = E .dA = q_{int} /ε₀
inside the spherical shell there are no charges
The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.
We have two sphere shells with radii 0.050m and 0.15m respectively
a) point where you want to know the electric field d = 0.20 m
shell 1
the point is on the outside,d>ro, therefore we can consider the charge to be concentrated in the center
E₁ = k q₁ / d²
shell 2
the point is on the outside,d>ro
E₂ = k q₂ / d²
the total camp is
E_total = -E₁ + E₂
E_total = k ( )
E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²
E_total = 6,525 10⁵ N /C
The field direction is radial and outgoing ti the shells
b) the calculation point is d = 0.10m
shell 1
point outside the shell d> ro
E₁ = k q₁ / d²
shell 2
the point is inside the shell d <ro
Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero
E₂ = 0
E_total = E₁
E_total = k q₁ / d²
E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²
E_total = 1.89 10⁶ N / A
As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1
c) the point of interest d = 0.025 m
shell 1
point is inside the shell d< ro
as there are no charges inside
E₁ = 0
shell 2
point is inside the radius of the shell d <ro
E₂ = 0
the total field is
E_total = 0