<em>HERE'S YOUR ANSWER...</em>
• In coastal areas have less variation in temperature that. non-coastal area due to sea-breeze and land-breeze.
<em>EXPLANATION:</em>
• In day time the land heats up faster than the sea, so air above sand gets heated and rises up and cool air above ocean moves towards land to take its place.
• At night the land cools down faster than sea which causes hot air above sea to rise up and cool air above land moves in to take its place.
<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPS</em><em>.</em><em>.</em><em>.</em>
Had to look for the options and here is my answer.
Based on the listed choices, the one that is considered as a disadvantage of radio telescopes over optical telescopes is that "t<span>hey have greater resolution for similar size collectors." Hope this answers your question.</span>
Quasi frequency = 4√6
Quasi period = π√6/12
t ≈ 0.4045
<u>Explanation:</u>
Given:
Mass, m = 20g
τ = 400 dyn.s/cm
k = 3920
u(0) = 2
u'(0) = 0
General differential equation:
mu" + τu' + ku = 0
Replacing the variables with the known value:
20u" + 400u' + 3920u = 0
Divide each side by 20
u" + 20u' + 196u = 0
Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.
r² + 20r + 196 = 0
Determining the roots:
r = -10 ± 4√6i
The general solution for two complex roots are:
y = c₁ eᵃt cosbt + c₂ eᵃt sinbt
with a the real part of the roots and b be the imaginary part of the roots.
Since, a = -10 and b = 4√6
u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t
u(0) = 2
u'(0) = 0
(b)
Quasi frequency:
μ =
(c)
Quasi period:
T = 2π / μ
(d)
|u(t)| < 0.05 cm
u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05
solving for t:
τ = t ≈ 0.4045
Using conservation of momentum, we can solve for the force that the air bag exerts on the person.
Recall the equation for momentum (p):
We can solve for total momentum, then divide by out time interval. This gets us:
F = 4800N
Answer: option B. 45°
Explanation:
A body thrown at an angle from horizontal undergoes a projectile motion. The maximum horizontal displacement means the maximum range covered. The range is given by:
where v is the initial velocity, is angle of the projectile and g is the acceleration due to gravity.
For maximum range, sin 2θ = 1
2θ = sin⁻¹ 1 = 90°
θ = 90°/2 = 45°
Thus, the projectile undergoes maximum horizontal displacement at angle of 45°. Correct option is B.