Answer:
<h2>C. </h2>
Explanation:
<h3>#CARRY ON LEARNING</h3><h3>#MARK ON LEARNING</h3><h3>#HELPING HAND</h3>
Explanation:
F net = 2+6-4 ( 2 and 6 N are in same direction so they get added, 4N in opposite direction so it will be subtracted)
F net=4 N
Answer:
Sound wave X amplitude is greater than 'A' and its frequency is lesser than
'f'
Explanation:
The pitch of a sound is dictated by the frequency of the sound wave, while the loudness is dictated by the amplitude.
A high pitch sound corresponds to a high frequency and a low pitch sound corresponds to a low frequency.
The larger the amplitude of the waves, the louder the sound and vice-versa.
From the question,
Sound wave W has amplitude ‘A' and frequency 'f' and
Sound wave X is louder and lower in pitch than sound wave W.
Since sound wave X is louder, this means its amplitude is greater than 'A'.
Also, since sound wave X is lower in pitch, this means its frequency is lesser than 'f'.
Answer:
(a)-0.701m
(b)-0.674m/s
Explanation:
Given a = -0.24,m/s² Xo = 0.170m, u = 0.190m/s
X = Xo + ut +1/2at²
X = 0.170 + 0.190×3.6 + 1/2×(-0.24) 3.6² = -0.701m
(b) its velocity at the end of this time interval is v
v = u + at
= 0.190 + (-0.24)× 3.6 = -0.674m/s
Answer:
The answer to the question is
The luminosity of stars A is four times that of star B
Explanation:
Flux (F) produced by a source of light is directly proportional to the brightness or Luminosity (L), and varies inversely to its distance d, that is 
Therefore if the two stars present the same flux then we have
then crossing out like terms gives
or 4·L₁ = L₂
The luminosity of star A is 4 times the that of star B