Answer:
2.8 cm
Explanation:
= Separation between two first order diffraction minima = 1.4 cm
D = Distance of screen = 1.2 m
m = Order
Fringe width is given by
![\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm](https://tex.z-dn.net/?f=%5Cbeta_1%3D%5Cdfrac%7By_1%7D%7B2%7D%5C%5C%5CRightarrow%20%5Cbeta_1%3D%5Cdfrac%7B1.4%7D%7B2%7D%5C%5C%5CRightarrow%20%5Cbeta_1%3D0.7%5C%20cm)
Fringe width is also given by
![\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}](https://tex.z-dn.net/?f=%5Cbeta_1%3D%5Cdfrac%7Bm_1%5Clambda%20D%7D%7Bd%7D%5C%5C%5CRightarrow%20d%3D%5Cdfrac%7Bm_1%5Clambda%20D%7D%7B%5Cbeta_1%7D)
For second order
![\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1](https://tex.z-dn.net/?f=%5Cbeta_2%3D%5Cdfrac%7Bm_2%5Clambda%20D%7D%7Bd%7D%5C%5C%5CRightarrow%20%5Cbeta_2%3D%5Cdfrac%7Bm_2%5Clambda%20D%7D%7B%5Cdfrac%7Bm_1%5Clambda%20D%7D%7B%5Cbeta_1%7D%7D%5C%5C%5CRightarrow%20%5Cbeta_2%3D%5Cdfrac%7Bm_2%7D%7Bm_1%7D%5Cbeta_1)
Distance between two second order minima is given by
![y_2=2\beta_2](https://tex.z-dn.net/?f=y_2%3D2%5Cbeta_2)
![\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%20y_2%3D2%5Cdfrac%7Bm_2%7D%7Bm_1%7D%5Cbeta_1%5C%5C%5CRightarrow%20y_2%3D2%5Cdfrac%7B2%7D%7B1%7D%5Ctimes%200.7%5C%5C%5CRightarrow%20y_2%3D2.8%5C%20cm)
The distance between the two second order minima is 2.8 cm
Answer:
1 hour to ride his motorcycle
Answer:
1.43 s
Explanation:
The time it takes for the container to reach the ground is determined only by the vertical motion of the container, which is a free-fall motion, so a uniformly accelerated motion with a constant acceleration of g=9.8 m/s^2 towards the ground.
The vertical distance covered by an object in free fall is given by
![S=ut + \frac{1}{2}at^2](https://tex.z-dn.net/?f=S%3Dut%20%2B%20%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where
u = 0 is the initial vertical speed
t is the time
a= g = 9.8 m/s^2 is the acceleration
since u=0, it can be rewritten as
![S=\frac{1}{2}gt^2](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
And substituting S=10.0 m, we can solve for t, to find the duration of the fall:
![t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2S%7D%7Bg%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%2810.0%20m%29%7D%7B9.8%20m%2Fs%5E2%7D%7D%3D1.43%20s)
Answer:C) car X
Explanation:
Given
All the cars have identical Engine thus Force Produced by car X will be equal to Y and Z
and ![Force=mass\times acceleration](https://tex.z-dn.net/?f=Force%3Dmass%5Ctimes%20acceleration)
Since Car X is most massive so acceleration associated with it will be minimum
acceleration of car X is minimum thus it will travel farthest
The meter out circuit is the flow control circuit design that can most effectively control an overrunning load.
The meter-out circuit can be very accurate, but are not efficient. The meter-out circuit can control overrunning as well as opposing loads while the other one method must be used with opposing loads only. The choice of flown control valve method and the location of the flow control in the circuit are dependent on the type of application being controlled.
<h3>What is a Circuit ?</h3>
In electronics, a circuit is a complete circular conduit through which electricity flows. A simple circuit consists of conductors, a load, and a current source. The term "circuit" broadly refers to any continuous path via which electricity, data, or a signal might flow.
- The directional valve shifts, causing the actuator to move faster than pump flow can fill it due to an overrunning load. Oil is leaking from one side, whereas there is none on the other.
Hence, flow control circuit design that can best control an overrunning load is the opposing circuit
Learn more about Circuit here:
brainly.com/question/26064065
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