Ox:vₓ=v₀
x=v₀t
Oy:y=h-gt²/2
|vy|=gt
tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g
y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°
cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s
Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ
Answer:
v= 1.71 m/s
Explanation:
Given that
Distance between two successive crests = 4.0 m
λ = 4 m
T= 7 sec
T is the time between 3 waves.
3 waves = 7 sec
1 wave = 7 /3 sec
So t= 7/3 s
We know that frequency f
f= 1/t= 3/7 Hz
Lets take speed of the wave is v
v= f λ
f=frequency
λ=wavelength
v= 3/7 x 4 = 12 /7
v= 1.71 m/s
Answer:
i) 3.514 s, ii) 5.692 m/s
Explanation:
i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.
as the equation says for free-falling
h = ut +0.5gt^2
Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)
initial velocity u = 0
10 = 0.5×1.62t^2
t = 3.514 seconds
Therefore, it takes t = 3.514 seconds for the Eagle to touch down.
ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon
v = u + gt
v = 0+ 1.62×3.514
v= 5.692 m/s
