Answer:
54.6°
Explanation:
From law of reflection i=r.
So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.
Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.
If you consider triangle AOB, one angle is ∠AOB=90°
and ∠OAB is 54.6°
From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°
So, the second incident angle will be 54.6°
Hence, the second reflected angle will be 54.6 degrees.
Electroreception is limited to aquatic environments because on here is the resistivity of the medium is low enough for electric currents to be generated as the result of electric fields of biological origin. In air, the resistivity of the environment is so high that electric fields from biological sources do not generate a significant electric current. Electroreceptor are found in a number of species of fish, and in at least one species of mammal, the Duck-Billed platypus.
Answer:
The work done by the child as the tricycle travels down the incline is 416.96 J
Explanation:
Given;
initial velocity of the child,
= 1.4 m/s
final velocity of the child,
= 6.5 m/s
initial height of the inclined plane, h = 2.25 m
length of the inclined plane, L = 12.4 m
total mass, m = 48 kg
frictional force,
= 41 N
The work done by the child is calculated as;

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J
Answer:
The tangential speed at Livermore is approximately 284.001 meters per second.
Explanation:
Let suppose that the Earth rotates at constant speed, the tangential speed (
), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:
(1)
Where:
- Rotation time, measured in seconds.
- Radius of the Earth, measured in meters.
- Latitude of the city above the Equator, measured in sexagesimal degrees.
If we know that
,
and
, then the tangential speed at Livermore is:


The tangential speed at Livermore is approximately 284.001 meters per second.
Answer:
Explanation:
Sam mass=75kg
Height is 50m
20° frictionless slope
Horizontal force on Sam is 200N
According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.
Therefore
Wg - Ww =∆K.E
Note initial the body was at rest at top of the slope.
Then, ∆K.E is K.E(final) - K.E(initial)
K.E Is given as ½mv²
Since initial velocity is zero then, K.E(initial ) is zero
Therefore, ∆K.E=½mVf²
Wg is work done by gravity and it is given by using P.E formulas
Wg=mgh
Wg=75×9.8×50
Wg=36750J
Ww is work done by wind and it's is given by using formulae for work
Work=force × distance
Ww=horizontal force × horizontal distance
Using Trig.
TanX=opposite/adjacent
Tan20=h/x
x=h/tan20
x=50/tan20
x=137.37m
Then,
Ww=F×x
Ww=200×137.37
We=27474J
Now applying the formula
Wg - Ww =∆K.E
36750 - 27474 =½×75×Vf²
9276=37.5Vf²
Vf²=9275/37.5
Vf²= 247.36
Vf=√247.36
Vf=15.73m/s