Answer:
OD. The wave diffracted as it passed through the opening in the barrier.
Explanation:
A progressive wave (i.e waves in motion) has the capacity to bend around an obstacle on its path. This is one of the general properties of waves called diffraction. Others are: reflection, refraction, interference. Note that only transverse waves undergo polarization.
Diffraction of waves is the ability of waves to bend around an obstacle on its path during progression.
Thus, the bending of the part of waves as it passes through the barrier implies that the wave diffracted as it passed through the opening in the barrier.
Answer:13.548 minutes
Explanation:
Given Speed of car A is 65 mph
Speed of car B is 90 mph
distance between cars at noon is 100 miles
as car A leaves ahead of car B thus car A travels extra for an hour which is equals to 65 miles
now the effective distance between cars is 35 mph
Now both car travels towards each other and meet after t hr
therefore
where x is distance covered by car A in time t
solving we get
x=14.677 miles
35-x=20.323 miles
they will pass each other after
Answer:
The acceleration is given by de second derivative of x(t) which is equal to m/s^2
Explanation:
a) We have the equation x(t)=at^4+bt^3+ct which is the position of the body of mass m at a time t
Where a, b and c are constants
From the rules of differenciation we have that the first derivative of the position is the velocity and the second derivative is the acceleration.
Hence the first derivative of the function is equal to [/tex] m/s
Don´t forget to write down the unities
Then we have to derivate again this equation, so we have
m/s^2[/tex]
b) Remembering the Newton´s laws we know that
where:
F is the force
m is the mass
and a is the acceleration
From the first part we know the value of the acceleration which is
m/s^2
So using the second law formula and replacing the values we have that
F=m( ) N
Remember the that N= Newton which is kg*m/s^2
Answer: 7.3seconds
Explanation:
Using the echo concept, before echo can occur, we must have an object making the sound (the teacher), the diatance between the object and the reflector (x)(i.e the wall), the time taken for the shadow to disappear (t) and the velocity of sound /movement of the object (v). Using the relationship 2x = vt
According to the question, x = 12m (initial distance), v = speed of object t = time of disappearance.
We have time before disappearance calculated as 2 × 12/1.1 = 21.8s
When he is 4m from the building, it will there take 4 × 21.8/12 = 7.3seconds for his shadow to decrease on the way.
Nothing will happen. The mercury and steel will not move. It’s just like filling a glass halfway with cement and then pouring water on top of it.