Answer:
How does the drink content affect an individual's blood pressure?
Explanation:
In every experiment using the scientific method, an observation lays the foundation of that experiment. A problem must be observed, which then leads to asking a SCIENTIFIC QUESTION in order to investigate. A scientific question must include the variable being changed called INDEPENDENT VARIABLE and the variable being measured called DEPENDENT VARIABLE.
In this experimental procedure or set up,
- Group 1 drinks 500 mL of coffee a day.
- Group 2 drink 500 mL of tea a day,
- Group 3 is a control group i.e no drink
At the end of 60 days all participants
blood pressure is tested.
This set up indicates that the variable being changed (independent) is the DRINK CONTENT while the variable being measured (dependent) is the BLOOD PRESSURE. Hence, these variables serve as the template to ask a scientific question which goes thus:
HOW DOES THE DRINK CONTENT AFFECT AN INDIVIDUAL'S BLOOD PRESSURE?
This scientific question relates how the independent variable (drink) causes the dependent variable to respond (blood pressure).
Answer:
frequency is 195.467 Hz
Explanation:
given data
length L = 4.36 m
mass m = 222 g = 0.222 kg
tension T = 60 N
amplitude A = 6.43 mm = 6.43 ×
m
power P = 54 W
to find out
frequency f
solution
first we find here density of string that is
density ( μ )= m/L ................1
μ = 0.222 / 4.36
density μ is 0.050 kg/m
and speed of travelling wave
speed v = √(T/μ) ...............2
speed v = √(60/0.050)
speed v = 34.64 m/s
and we find wavelength by power that is
power = μ×A²×ω²×v / 2 ....................3
here ω is wavelength put value
54 = ( 0.050 ×(6.43 ×
)²×ω²× 34.64 ) / 2
0.050 ×(6.43 ×
)²×ω²× 34.64 = 108
ω² = 108 / 7.160 ×
ω = 1228.16 rad/s
so frequency will be
frequency = ω / 2π
frequency = 1228.16 / 2π
frequency is 195.467 Hz
When driver see the child standing on road his speed is 20 m/s
So here at that instant his reaction time is 0.80 s
He will cover a total distance given by product of speed and time
![d_1 = v* t](https://tex.z-dn.net/?f=d_1%20%3D%20v%2A%20t)
![d_1 = 20 * 0.8](https://tex.z-dn.net/?f=d_1%20%3D%2020%20%2A%200.8)
![d_1 = 16 m](https://tex.z-dn.net/?f=d_1%20%3D%2016%20m)
now after this he will apply brakes with acceleration a = 7 m/s^2
so the distance covered before it stop is given by
![v_f^2 - v_i^2 = 2 a d](https://tex.z-dn.net/?f=v_f%5E2%20-%20v_i%5E2%20%3D%202%20a%20d)
![0 - 20^2 = 2*(-7)*d_2](https://tex.z-dn.net/?f=0%20-%2020%5E2%20%3D%202%2A%28-7%29%2Ad_2)
![d_2 = 28.6 m](https://tex.z-dn.net/?f=d_2%20%3D%2028.6%20m)
so the total distance covered by it
![d = d_1 + d_2](https://tex.z-dn.net/?f=d%20%3D%20d_1%20%2B%20d_2)
![d = 16 + 28.6 = 44.6 m](https://tex.z-dn.net/?f=d%20%3D%2016%20%2B%2028.6%20%3D%2044.6%20m)
<em>so it will cover a total distance of 44.6 m</em>
Explanation:
Given that,
Angle by the normal to the slip α= 60°
Angle by the slip direction with the tensile axis β= 35°
Shear stress = 6.2 MPa
Applied stress = 12 MPa
We need to calculate the shear stress applied at the slip plane
Using formula of shear stress
![\tau=\sigma\cos\alpha\cos\beta](https://tex.z-dn.net/?f=%5Ctau%3D%5Csigma%5Ccos%5Calpha%5Ccos%5Cbeta)
Put the value into the formula
![\tau=12\cos60\times\cos35](https://tex.z-dn.net/?f=%5Ctau%3D12%5Ccos60%5Ctimes%5Ccos35)
![\tau=4.91\ MPa](https://tex.z-dn.net/?f=%5Ctau%3D4.91%5C%20MPa)
Since, the shear stress applied at the slip plane is less than the critical resolved shear stress
So, The crystal will not yield.
Now, We need to calculate the applied stress necessary for the crystal to yield
Using formula of stress
![\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}](https://tex.z-dn.net/?f=%5Csigma%3D%5Cdfrac%7B%5Ctau_%7Bc%7D%7D%7B%5Ccos%5Calpha%5Ccos%5Cbeta%7D)
Put the value into the formula
![\sigma=\dfrac{6.2}{\cos60\cos35}](https://tex.z-dn.net/?f=%5Csigma%3D%5Cdfrac%7B6.2%7D%7B%5Ccos60%5Ccos35%7D)
![\sigma=15.13\ MPa](https://tex.z-dn.net/?f=%5Csigma%3D15.13%5C%20MPa)
Hence, This is the required solution.