Answer:
a = 1 m/s² and
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
Ty - W = 0
Ty = w
X axis
Tx = m a
With trigonometry we find the components of tension
Sin θ = Ty / T
Ty = T sin θ
Cos θ = Tx / T
Tx = T cos θ
We calculate the acceleration with kinematics
Vf = Vo + a t
a = (Vf -Vo) / t
a = (20 -10) / 10
a = 1 m/s²
We substitute in Newton's equations
T Sin θ = mg
T cos θ = ma
We divide the two equations
Tan θ = g / a
θ = tan⁻¹ (g / a)
θ = tan⁻¹ (9.8 / 1)
θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle
The orbital radius is: 
Explanation:
The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.
For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:
where
G is the gravitational constant
M is the mass of the planet
m is the mass of the satellite
r is the radius of the orbit
v is the speed of the satellite
Re-arranging the equation, we find:
Learn more about circular motion:
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Answer:
Low Potential energy and High Kinetic energy
Explanation:
Hope this helps and have a good day! Apologies if it's wrong.<3
Answer:
the magnetic field is leaving the sheet
Explanation:
The magnetic force is given by the expression
F = q v x B
where bold letters indicate vectors, the modulus of this expression is
F = q v B sin θ
the direction of the force is given by the right hand rule, for a positive charge
the thumb indicates the direction of the speed, in this case from right to left
the palm the direction of the force, in our case upwards
the fingers extended the direction of the magnetic field, this case after fixing the other two components it points out of the blade
In short the magnetic field is leaving the sheet
Answer:
λ = 162 10⁻⁷ m
Explanation:
Bohr's model for the hydrogen atom gives energy by the equation
= - k²e² / 2m (1 / n²)
Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer
The Planck equation
E = h f
The speed of light is
c = λ f
E = h c /λ
For a transition between two states we have
- 
= - k²e² / 2m (1 / 
² -1 / 
²)
h c / λ = -k² e² / 2m (1 / ² - 1/ ²)
1 / λ = (- k² e² / 2m h c) (1 / ² - 1/²)
The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses
Let's calculate the emission of the transition
1 /λ = 1.097 10⁷ (1/10² - 1/8²)
1 / λ = 1.097 10⁷ (0.01 - 0.015625)
1 /λ = 0.006170625 10⁷
λ = 162 10⁻⁷ m
