Answer:
a)5m b)5secs c) 10secs
Explanation:
a) maximum height H = u²/2g
Given u = 50m/s
g = 10m/s²
H = 10²/20
H= 5m
maximum height reached by the rocket is therefore 5m
.b)Time it takes the rocket to reach its maximum height is Tmax = U/g
Tmax = 50/10
Tmax = 5seconds
c) How long the rocket spent in the air is the time of flight T
T = 2U/g
T = 2×50/10
T = 100/10
T = 10seconds
Answer:
Yes, correct
Explanation:
velocity, v = 470 m/s
radius, r = 0.15 m
The radial acceleration is the centripetal acceleration which always acts towards the centre of the circular centrifuge.
The formula for the centripetal acceleration is given by
![a =\frac{v^{2}}{r}](https://tex.z-dn.net/?f=a%20%3D%5Cfrac%7Bv%5E%7B2%7D%7D%7Br%7D)
![a =\frac{470^{2}}{0.15}](https://tex.z-dn.net/?f=a%20%3D%5Cfrac%7B470%5E%7B2%7D%7D%7B0.15%7D)
a = 1472666.667
a = 150272.1 g
According to the question, we can get the acceleration as mentioned. So the claim is correct.
Answer:
Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s
Explanation:
Given Data:
V is the overall velocity vector,
and
are its initial vertical and horizontal components
![R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi = 10*sin(30) = 5 m/s\\Ui = 10*cos(30) = 8.66 m/s](https://tex.z-dn.net/?f=R%20%3D%2010%20m%2Fs%5C%5C%20Projection%20Angle%20%28theta%29%20%3D%2030%20degrees%5C%5CVi%20%20%20%3D%2010%2Asin%2830%29%20%3D%205%20m%2Fs%5C%5CUi%20%20%3D%2010%2Acos%2830%29%20%3D%208.66%20m%2Fs)
To find:
Max Height
achieved
Calculation:
1) Using the
equation of motion, we know
![2*a*s = Vf^{2} - Vi^{2}](https://tex.z-dn.net/?f=2%2Aa%2As%20%3D%20Vf%5E%7B2%7D%20%20-%20Vi%5E%7B2%7D)
2) In terms of gravity
height
and the vertical component of Velocity
.
3) As
as at maximum height the vertical component of velocity is zero maximum height achieved
![2*g*h = Vf^{2} -Vi^{2}](https://tex.z-dn.net/?f=2%2Ag%2Ah%20%3D%20Vf%5E%7B2%7D%20%20-Vi%5E%7B2%7D)
putting values
4) ![h = 0.784 m/s](https://tex.z-dn.net/?f=h%20%3D%200.784%20m%2Fs)
5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)
Answer:
25100A
Explanation:
t= 1ms = 0.001s
q = 25.1C
From the relationship between charge and current , the charge is equal to the product of current and time
q = i×t
Where q = charge
i = current
t = time
i = q/t = 25.1/0.001
i = 25100A.
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