Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.
![F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }](https://tex.z-dn.net/?f=F_%7Bn%7D%20%3D%5Csqrt%7B%28F_%7Bnx%7D%29%5E%7B2%7D%20%2B%28F_%7Bny%7D%29%20%5E%7B2%7D%20%7D)
![F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }](https://tex.z-dn.net/?f=F_%7Bn%7D%20%3D%5Csqrt%7B%28-7.62%29%5E%7B2%7D%20%2B%2829.13%29%20%5E%7B2%7D%20%7D)
![F_{n} = 30.11N](https://tex.z-dn.net/?f=F_%7Bn%7D%20%3D%2030.11N)
Direction of the net force (β)
![\beta =tan^{-1} (\frac{29.13}{7.62} )](https://tex.z-dn.net/?f=%5Cbeta%20%3Dtan%5E%7B-1%7D%20%28%5Cfrac%7B29.13%7D%7B7.62%7D%20%29)
β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
The answer is A. The Sun and all the planets revolve around Earth.
Aristotle believed that the Earth was the centre of the solar system, and the Sun and the planets orbited around it. He believed that the universe was composed of Earth-like bodies, which were at rest, and of heavenly bodies, which were in perpetual motion.
Answer:
70.07 Hz
Explanation:
Since the sound is moving away from the observer then
and
when moving towards observer
With
of 76 then taking speed in air as 343 m/s we have
![76 = f_s\times\frac {(343-vs)}{343}](https://tex.z-dn.net/?f=76%20%3D%20f_s%5Ctimes%5Cfrac%20%7B%28343-vs%29%7D%7B343%7D)
![f_s=\frac {343\times 76}{343-v_s}](https://tex.z-dn.net/?f=f_s%3D%5Cfrac%20%7B343%5Ctimes%2076%7D%7B343-v_s%7D)
Similarly, with
of 65 we have
![65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}](https://tex.z-dn.net/?f=65%20%3D%20f_s%5Ctimes%5Cfrac%20%7B%28343%2Bvs%29%7D%7B343%7D%5C%5Cf_s%3D%5Cfrac%20%7B343%5Ctimes%2065%7D%7B343%2Bv_s%7D)
Now
![f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}](https://tex.z-dn.net/?f=f_s%3D%5Cfrac%20%7B343%5Ctimes%2065%7D%7B343%2Bv_s%7D%3D%5Cfrac%20%7B343%5Ctimes%2076%7D%7B343-v_s%7D)
v_s=27.76 m/s
Substituting the above into any of the first two equations then we obtain
![f_s=70.07 Hz](https://tex.z-dn.net/?f=f_s%3D70.07%20Hz)
For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.