Answer:0.2mm
Explanation:
The length of one VSD=8/10=0.8mm
The least count of the instrument is the difference between the length of one MSD and length of one VSD
The length of inebriated MSD=1mm
Therefore,
The least count=1-0.8=0.2mm
Answer:
1)
2)
Explanation:
<u>Projectile Motion</u>
When an object is launched near the Earth's surface forming an angle 
with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.
The heigh of an object can be computed as
Where 
is the initial height above the ground level, 
is the vertical component of the initial velocity and t is the time
The y-component of the speed is
1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of 
The object will reach the maximum height when 
. It allows us to compute the time to reach that point
Solving for 
Thus, the maximum heigh is
We know this value is 8 meters
Solving for
Replacing the known values
2) We know at t=1.505 sec the ball is above Julie's head, we can compute
Answer:
a) W = 643.5 J, b) W = -427.4 J
Explanation:
a) Work is defined by
W = F. x = F x cos θ
in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero
W = F x
let's calculate
W = 165 3.9
W = 643.5 J
b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes
sin θ = Wₓ / W
cos θ = Wy / W
Wₓ = W sinθ = mg sin θ
Wy = W cos θ
the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero
W = Wₓ x cos 180
W = - mg sin 34 x
let's calculate
W = -20 9.8 sin 34 3.9
W = -427.4 J
The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.
Answer:R=1607556m
θ=180degrees
Explanation:
d1=74.8m
d2=160.7km=160.7km*1000
d2=160700m
d3=80m
d4=198.1m
Using analytical method :
Rx=-(160700+75*cos(41.8))= -160755.9m
Ry= -(74.8+75sin(41.8))-198.1=73m
Magnitude, R:
R=√Rx+Ry
R=√160755.9^2+20^2=160755.916
R=160756m
Direction,θ:
θ=arctan(Rx/Ry)
θ=arctan(-73/160755.9)
θ=-7.9256*10^-6
Note that θ is in the second quadrant, so add 180
θ=180-7.9256*10^6=180degrees
When air is blown into the open pipe,
L = 
where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation
⇒λ=
Note here that n=1 is for fundamental, n=2 is first harmonic and so on..
⇒ third harmonic will be n=4
Given L=6m, n=4, solving for λ we get:
λ=
=3m
Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:
c=f.λ Or f= 
⇒f=
≈115 Hz
