Answer:
(a) V = 0.75 m/s
(b) V = 0.125 m/s
Explanation:
The speed of the flow of the river can be given by following formula:
V = Q/A
V = Q/w d
where,
V = Speed of Flow of River
Q = Volume Flow Rate of River
w = width of river
d = depth of river
A = Area of Cross-Section of River = w d
(a)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 20 m
d = 20 m
Therefore,
V = (300 m³/s)/(20 m)(20 m)
<u>V = 0.75 m/s</u>
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(b)
Here,
Q = (300,000 L/s)(0.001 m³/1 L) = 300 m³/s
w = 60 m
d = 40 m
Therefore,
V = (300 m³/s)/(60 m)(40 m)
<u>V = 0.125 m/s</u>
The spring force = kx = mg.
k is spring constant = 200 N/m
x is the spring elongation = 0.2 m
F = 200×0.2= 40 N
Now, F = mg wil be the weight of the object
Thus, weight is 40 N.
Thus, now we can also find the mass of object, m = F/g = 40 N/9.8
Because g is acceleration due to gravity = 9.8 m/sec ^2
m = 4.08 kg
Answer:
0.71 m
Explanation:
Let the 5 kg is hang at a distance d from the midpoint.
use the concept of moments.
take moments about the mid point.
Anticlockwise moments = clockwise moments
3 x (0.5 - 0.15) = 5 x d
1.05 = 5 d
d = 0.21 m
So, the distance of 5 kg from 0 mark on the stick is 0.5 + 0.21 = 0.71 m
Answer:
Distance, s = 399.42 m/s
Explanation:
It is given that,
Initial velocity of drag racer, u = 0
Final velocity of drag racer, v = 407 km/hr = 113.056 m/s
Acceleration, ![a=16\ m/s^2](https://tex.z-dn.net/?f=a%3D16%5C%20m%2Fs%5E2)
Third equation of motion is :
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
s is the distance traveled
![s=\dfrac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
![s=\dfrac{v^2}{2a}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7Bv%5E2%7D%7B2a%7D)
![s=\dfrac{(113.056)^2}{2\times 16}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7B%28113.056%29%5E2%7D%7B2%5Ctimes%2016%7D)
s = 399.42 m/s
So, the distance traveled during this acceleration is 399.42 m/s. Hence, this is the required solution.
Answer:
1.15 rad/s²
Explanation:
given,
angular speed of turntable = 45 rpm
=![45\times \dfrac{2\pi}{60}](https://tex.z-dn.net/?f=45%5Ctimes%20%5Cdfrac%7B2%5Cpi%7D%7B60%7D)
=![4.71\ rad/s](https://tex.z-dn.net/?f=4.71%5C%20rad%2Fs)
time, t = 4.10 s
initial angular speed = 0 rad/s
angular acceleration.
![\alpha = \dfrac{\omega_f-\omega_0}{t}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cdfrac%7B%5Comega_f-%5Comega_0%7D%7Bt%7D)
![\alpha = \dfrac{4.71-0}{4.10}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cdfrac%7B4.71-0%7D%7B4.10%7D)
![\alpha = 1.15\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%201.15%5C%20rad%2Fs%5E2)
Hence, the angular acceleration of the turntable is 1.15 rad/s²