Force = Mass . Accelaration
f=ma
force is directly proportional to mass and accelaration
therefore when mass of the object is doubled, the accelaration would be halved
since
a=f/m
<span>40.7 miles.
For this problem, we want to know the length of the chord created by the line and the circle. So let's first create the equations needed.
The slope intercept equation for a line is:
y = ax + b
the value for a will be the the difference in y divided by the difference in x. We're going from y=61 to y=0 for a chance of -61 and from x=0 to x=62 for a change of 62. So the value of a is
-61/62, giving us the formula
y = -(61/62)x + b
Substituting x = 0, we can calculate b
61 = -(61/62)0 + b
61 = b
So the equation for the line is:
y = -(61/62)x + 61
Now for the equation for the circle. Since the circle is centered at the origin, the equation is:
x^2 + y^2 = 48^2
Now we to calculate the intersections.
y = -(61/62)x + 61
x^2 + y^2 = 48^2
x^2 + (-(61/62)x + 61)^2 = 48^2
x^2 + (3721/3844)x^2 - (3721/31)x + 3721 = 2304
(7565/3844)x^2 - (3721/31)x + 3721 = 2304
(7565/3844)x^2 - (3721/31)x + 1417 = 0
1.968002081x^2 - 120.0322581x + 1417 = 0
And we have a rather ugly quadratic equation which we can solve using the quadratic formula, giving the solutions x = 16.00512574 and x = 44.98681081
Now we need to calculate the y values for those 2 x values.
y = -(61/62)x + 61
y = -(61/62)16.00512574 + 61
y = 45.25302145
y = -(61/62)x + 61
y = -(61/62)44.98681081 + 61
y = 16.73878292
So the 2 endpoints are
(16.00512574, 45.25302145) and (44.98681081, 16.73878292)
The distance between those points can be calculated using the Pythagorean theorem.
sqrt((16.00512574 - 44.98681081)^2 + (45.25302145 - 16.73878292)^2)
= sqrt(-28.98168506^2 + 28.51423853^2)
=
sqrt(839.938069 + 813.0617988)
=
sqrt(1652.999868)
= 40.65710107
And finally, we have the solution of 40.7 miles.</span>
Answer:
Explanation:
Given
Charge on particles
![q_1=+9\ nC](https://tex.z-dn.net/?f=q_1%3D%2B9%5C%20nC)
![q_2=-3\ nC](https://tex.z-dn.net/?f=q_2%3D-3%5C%20nC)
![x_1=-2\ cm](https://tex.z-dn.net/?f=x_1%3D-2%5C%20cm)
![x_2=13\ cm](https://tex.z-dn.net/?f=x_2%3D13%5C%20cm)
Third charge
must be placed right side of
as
will attract and
so net force will be zero
Electrostatic force is given by
![F=\frac{kq_1q_2}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7Bkq_1q_2%7D%7Br%5E2%7D)
suppose
is at a distance of x cm from ![q_2](https://tex.z-dn.net/?f=q_2)
![F_{13}=\frac{kq_1q_3}{(13+2+x)^2}](https://tex.z-dn.net/?f=F_%7B13%7D%3D%5Cfrac%7Bkq_1q_3%7D%7B%2813%2B2%2Bx%29%5E2%7D)
![F_{23}=\frac{kq_2q_3}{(x)^2}](https://tex.z-dn.net/?f=F_%7B23%7D%3D%5Cfrac%7Bkq_2q_3%7D%7B%28x%29%5E2%7D)
![F_{13}+F_{23}=0](https://tex.z-dn.net/?f=F_%7B13%7D%2BF_%7B23%7D%3D0)
![\frac{k(9)(q_3)}{(15+x)^2}=\frac{k3q_3}{(x)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bk%289%29%28q_3%29%7D%7B%2815%2Bx%29%5E2%7D%3D%5Cfrac%7Bk3q_3%7D%7B%28x%29%5E2%7D)
![x(\sqrt{3}-1)=15](https://tex.z-dn.net/?f=x%28%5Csqrt%7B3%7D-1%29%3D15)
![x=\frac{15}{\sqrt{3}-1}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B15%7D%7B%5Csqrt%7B3%7D-1%7D)
![x=20.49\ cm](https://tex.z-dn.net/?f=x%3D20.49%5C%20cm)
<h2>
Answer: a.The mirrors and eyepiece of a large telescope are spring-loaded to allow them to return quickly to a known position. </h2>
Explanation:
Adaptive optics is a method used in several astronomical observatories to counteract in real time the effects of the Earth's atmosphere on the formation of astronomical images.
This is done through the insertion into the optical path of the telescope of sophisticated deformable mirrors supported by a set of computationally controlled actuators. Thus obtaining clear images despite the effects of atmospheric turbulence that cause the unwanted distortion.
It should be noted that with this technique it is also necessary to have a moderately bright reference star that is very close to the object to be observed and studied. However, it is not always possible to find such stars, so a powerful laser beam is used to point towards the Earth's upper atmosphere and create artificial stars.
The correct answer is B. hopes this help