F=1/T
therefore T=1/F
so thats T=1/20x10^6
T=5x10^-8seconds
Answer: e. on the equator and in the Atlantic Ocean
Explanation:
Latitude and Longitude are geographical coordinates.
<u>Latitude</u> is the angular distance between the <u>equatorial line</u>, and a specific point on the Earth. It is measured in degrees and is represented according to the hemisphere in which the point is located, which can be north or south latitude.
In this sense latitude
refers to the equatorial line that divides the Earth in two hemispheres (North and South).
On the other hand, <u>Longitude</u> represents the specific <u>east–west </u>position of a point on the Earth's surface, being longitude
the prime meridian or Greenwich meridian.
So, if we have latitude
and longitude
(positive means it is to the East) we can already know the point is in the equator, and the option that bests describes the coordinates is e.
Answer:
ρ = 1469 kg/m³
Explanation:
given,
mass of statue = 0.4 Kg
density of statue = 8 x 10³ kg/m³
tension in the string = 3.2 N
density of the fluid = ?
Volume of the statue

V = 5 x 10⁻⁵ m³
W = ρ g V
W = ρ x 9.8 x 5 x 10⁻⁵
now, tension on the string will be equal to
T = mg - W
3.2 = 0.4 x 9.8 - ρ x 9.8 x 5 x 10⁻⁵
ρ x 9.8 x 5 x 10⁻⁵ = 0.72
ρ = 1469 kg/m³
Answer:
ee that the lens with the shortest focal length has a smaller object
Explanation:
For this exercise we use the constructor equation or Gaussian equation
where f is the focal length, p and q are the distance to the object and the image respectively.
Magnification a lens system is
m =
= -
h ’= -\frac{h q}{p}
In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞
Let's calculate the distance to the image for each lens
f = 6.0 cm

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf
q = f = 6.0 cm
for the lens of f = 12.0 cm q = 12.0 cn
to find the size of the image we use
h ’= h q / p
where p has a high value and is the same for all systems
h ’= h / p q
Thus
f = 6 cm h ’= fo 6 cm
f = 12 cm h ’= fo 12 cm
therefore we see that the lens with the shortest focal length has a smaller object