Answer:
a) "gravitation" is the force causing you to go down a waterslide
b) It is "fluid friction" as a solid object (our body) moves over a fluid (the water)
c) It would become "sliding friction" since two solid surfaces slide over each other
d) fluid friction being the weakest friction, switching to sliding friction means a higher decrease in speed and therefore removing the water from a slide will decrease our speed
Solving this using the time, we know that range = horizontal velocity x time of flight
since
there are no horizontal forces acting on the ball, there are no
horizontal accelerations and the initial horizontal velocity of 36 cos
28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have
range = 36 cos 28 x 3.44 s = 109.3 m
The work done by 
along the given path <em>C</em> from <em>A</em> to <em>B</em> is given by the line integral,
I assume the path itself is a line segment, which can be parameterized by
with 0 ≤ <em>t</em> ≤ 1. Then the work performed by <em>F</em> along <em>C</em> is
![\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Cleft%286x%28t%29%5E3%5C%2C%5Cvec%5Cimath-4y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%29%5Ccdot%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Bx%28t%29%5C%2C%5Cvec%5Cimath%20%2B%20y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%5D%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%20%28288%283t-1%29%5E3-8%282t%2B5%29%29%20%5C%2C%5Cmathrm%20dt%20%3D%20%5Cboxed%7B312%7D)
The calculated magnitude is 6.73 x 10³ V/m.
AMU is described as being one-twelfth the mass of a carbon-12 atom (12C). C makes up more than 98% of the carbon that can be found in nature, making it the most prevalent isotope. The magnitude of the field is the change in potential across a small distance in the indicated direction divided by that distance.
Potential difference = 8.20 kV= 8.20 x 10³ V
radius= 19.4/100=0.194 m
total distance that is circumference of the circle= 2πr =2 x 3.14 x 0.194
= 1.218 m
therefore Magnitude= 8.20 x 10³ / 1.218
=6.73 x 10³ V/m
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We have that the Number of stitches per sec and he mass of oscillation motion is mathematically given as
a) Nt=25stitches per sec
b) m=2.033e-5kg
<h3>
Number of
stitches per sec and he mass of oscillation motion</h3>
Question Parameters:
This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.
If the <em>sewing </em>machine has a spring constant of 0.5 N/m,
Generally the equation for the Number of stitches per sec is mathematically given as
Nt=N/t
Therefore
Nt=1500/60
Nt=25stitches per sec
b)
Generally the equation for the Time t is mathematically given as
Therefore
m=2.033e-5kg
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