Answer:Principle of rectilinear propagation of light
Explanation:Principle of rectilinear propagation of light
Rectilinear propagation of light refers to the propensity of light to travel along a straight line without any interference in its trajectory. ... It is because light travels along a straight line and leaves only the areas where the object interferes.
The final speed of an airplane is v = 92.95 m/s
The rate of change of position of an object in any direction is known as speed i.e. in other word, Speed is measured as the ratio of distance to the time in which the distance was covered.
Solution-
Here given,
Acceleration a= 10.8 m/s2 .
Displacement (s)= 400m
Then to find final speed of airplane v=?
Therefore from equation of motion can be written as,
v²=u²+ 2as
where, u is initial speed, v is final speed ,a is acceleration and s is displacement of the airplane. Therefore by putting the value of a & s in above equation and (u =0) i.e. the initial speed of airplane is zero.
v²= 2×10.8 m/s²×400m
v²=8640m/s
v=92.95m/s
hence the final speed of airplane v =92.95m/s
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Answer:
The initial velocity was U=22.14m/s
Explanation:
Step one :
Applying the third equation of motion
v² = u²+ 2as
Where v= Final velocity
U =initial velocity
a= acceleration due to gravity
S= distance or displacement
Step two :
V= 0
a= 9.81m/s²
S=25m
U=?
Step three :
Substituting into the equation we have
0²=U²+2*9.81*25
0=U²+490.5
U²=-490.5
U=√490.5
U=22.14m/s
Answer:
Ф,
Ф
Explanation:
Now find the components NxNxN_x and NyNyN_y of N⃗ N→N_vec in the tilted coordinate system of Part B. Express your answer in terms of the length of the vector NNN and the angle θθtheta, with the components separated by a comma.
Vectors are quantities that have both magnitude and direction while scalar quantities have only magnitude but no direction.
This a vector quantity
from the diagram the horizontal component of the length of the vector will be
Ф
the vertical component will be
Ф
this is in the opposite direction because the x can be extrapolated to the negative axis
Answer:
Torque,
Explanation:
Given that,
The loop is positioned at an angle of 30 degrees.
Current in the loop, I = 0.5 A
The magnitude of the magnetic field is 0.300 T, B = 0.3 T
We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

Let us assume that, 
is the angle between normal and the magnetic field, 
Torque is given by :

So, the net torque about the vertical axis is
. Hence, this is the required solution.