The angle is 53.5 degrees.
What is a flat mirror?
A plane mirror is a mirror with a flat (planar) reflective surface. For light rays striking a plane mirror, the angle of reflection equals the angle of incidence.
This ques have been solved in two steps:
Step 1:
given that;
Θ=107 degrees
from figure,
∠BAC=180-ψ-(90-φ)
∠BAC= 90-ψ-φ
∠AOC=180-2(ψ-φ)-2φ
∠AOC=180-2φ
Step 2: part(a)
writing an expression for the psi at point o:
Θ+180-2ψ=180
2ψ=Θ
ψ=Θ/2
above is the expression for ψ
Step 3: part(b)
The angle of ψ is,
ψ=Θ/2
ψ=107/2
ψ=53.5 degrees
Hence, the angle is 53.5 degrees.
To more know about flat mirrors and phi and psi angles the link is given below:
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Answer:
The magnitude of charge on each is 
Solution:
As per the question:
Mass of Earth, 
Mass of Moon, 
Now,
The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:
(1)
Now,
If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:
(2)
Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:
Equating eqn (1) and (2):
Q = 
Answer:
Explanation:
There is no external force applied, therefore there is a moment's preservation throughout the trajectory.
<em>Initial momentum = Final momentum. </em>
The total mass is equal to
Where,
mass of ship
mass of fuell expeled.
As the moment is conserved we have,
Where,
Velocity of fuel
Velocity of Space Ship
Solving and re-arrange to 
we have,
Where c is the speed of light.
Therefore the ship be moving with speed
Answer:
Δx = 2m
Explanation:
From the given information:
De la información dada:
El desplazamiento ocurrió en un vector que se mueve desde una posición inicial particular hasta el final.
Entonces, supongamos que el desplazamiento es D;
Luego:
D = (7,5 ) - (5,7)
D = (2, -2)
Por lo tanto, el valor del desplazamiento de la coordenada x; Δx = 2m
<span>So we want to know why the does a bouncing ball rise to a lower height with each bounce. So lets say the ball is first on some height h. There it has potential energy Ep=m*g*h. Then as the ball starts falling to the ground the energy converts to kinetic energy Ek=(1/2)*m*v^2. When the ball falls to the ground, the kinetic energy transforms to elastic energy because the ball deforms as it hits the ground and some small quantity of heat. The heat goes to the air and to the ground so it gets removed from the system. So there is less energy in the system to be converted back to kinetic energy as the ball starts to rise in height again. Thats why the ball is not able to get bact to the same height as it started from. </span>
