Answer:
Explanation:
The distance of searchlight will act as radius R and velocity of car may be supposed to be tangential velocity v . We are required to calculate angular velocity ω .
v = 80 ft /s
R = 3000 ft
ω = v / R
= 80 / 3000 = .027 rad / s
For angular acceleration the formula is
angular acceleration α = a / R
a is linear acceleration = 15 ft / s²
α = 15 / 3000 = .005 rad / s².
Explanation:
distance covered in the first part of journey : 120×14 = 1680km
distance covered in second part of journey :
90×10 = 900 km
total distance traveled : 1680+900 = 2580km.
total time taken to complete the journey : 14+10= 24 hrs.
therefore, the average speed of the journey = 2580km/24hr = 107.5 km/h.
hope this helps you.
Answer:
Option D is correct: 170 µW/m²
Explanation:
Given that,
Frequency f = 800kHz
Distance d = 2.7km = 2700m
Electric field Eo = 0.36V/m
Intensity of radio signal
The intensity of radial signal is given as
I = c•εo•Eo²/2
Where c is speed of light
c = 3×10^8m/s
εo = 8.85 × 10^-12 C²/Nm²
I = 3×10^8 × 8.85×10^-12 × 0.36²/2
I = 1.72 × 10^-4W/m²
I = 172 × 10^-6 W/m²
I = 172 µW/m²
Then, the intensity of the radio wave at that point is approximately 170 µW/m²
Answer:
1.Free edge
2.Nail Root
3.Nail matrix
4.lunula
Note: The numbering coincides respectively with the dashes in the questions.
Explanation:
Free edge is the distal white nail ending.
Nail Root is the proximal part of the nail embedded in the skin.
Nail matrix is the actively growing part of the nail were the nail root thickens
Lunula the whitish semilunar area of the proximal end of the nail body it appears whitish because stratum basala obscures. the underlying blood vessels.
