6mph:
Suppose the boat is traveling on a y axis. The river flow acts on the x axis. Motion on each axes are independent. The linear speed of the boat is not changed. Furthermore the projectile motion is changing, but you're specifically asking about the linear speed of the boat which is unchanged.
#1
for the block of mass 5 kg normal force is given as


friction force is given as


Net force is given as


now we know that



#2
Normal force is given as



now we know that


as object moves with constant velocity

now for coefficient of friction we can use



#3
net force upwards is given as

mass is given as

now as per newton's law we can say



#4
As we know that when block is sliding on rough surface
part a)
net force = applied force - frictional force




part b)
for coefficient of friction we can use


here normal force is given as

now we have

#5
if an object is initially at rest and moves 20 m in 5 s
so we can use kinematics to find out the acceleration



now net force is given as


#6
an object travelling with speed 25 m/s comes to stop in 1.5 s
so here acceleration of object is given as


now the force is gievn as


In position A
<em>The</em><em> </em><em>mirror </em><em>has</em><em> </em><em>to</em><em> </em><em>be</em><em> </em><em>observed</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>ray</em><em> </em><em>reflected</em><em> </em><em>from</em><em> </em><em>the</em><em> </em><em>mi</em><em>rror</em><em>.</em>
<em>Since</em><em> </em><em>the</em><em> </em><em>the</em><em> </em><em>angle</em><em> </em><em>of</em><em> </em><em>reflec</em><em>tion</em><em> </em><em>is</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>angle</em><em> </em><em>of</em><em> </em><em>incidence</em><em>,</em><em> </em><em>the</em><em> </em><em>reflected</em><em> </em><em>ray</em><em> </em><em>must</em><em> </em><em>be</em><em> </em><em>in</em><em> </em><em>equal</em><em> </em><em>path</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>incident</em><em> </em><em>ray</em><em>.</em>
<em>That</em><em> </em><em>phenomenon</em><em> </em><em>is</em><em> </em><em>only</em><em> </em><em>in</em><em> </em><em>diagram</em><em> </em><em>A</em>
Answer:
585 nm
Explanation:
The formula that gives the position of the m-th maximum (bright fringe) relative to the central maximum in the interference pattern produced by diffraction from double slit is:


where
m is the order of the maximum
is the wavelength
D is the distance of the screen from the slits
d is the separation between the slits
The distance between two consecutive bright fringes therefore is given by:

In this problem we have:
(distance between two bright fringes)
D = 2.0 m (distance of the screen)
d = 3.0 x 10−3 m (separation between the slits)
Solving for
, we find the wavelength:

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