<span>Star a is more distant and is approximately 5 times as far away as star b
Parallax is the change in angle that one must do in order to observe the same object from different locations. The further away an object is, the smaller the parallax is. As the angles approach zero, the trig functions tend to be fairly linear. And 0.1 arc seconds and 0.02 arc seconds are close enough to zero for this to hold true.
Since the parallax for star a is smaller than the parallax for star b, it is the more distant star. And since 0.1 divided by 0.02 = 5, it is approximately 5 times further away than star b.</span>
De los reptiles, anfibios; pájaros, peces; mamíferos y artrópodos, los que más se asemejan en características son los pájaros y los mamíferos porque ambos son de sangre caliente y no varia independientemente de en qué ambiente se encuentre
Answer:
Option (e)
Explanation:
A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,
Energy density = 100 J/m
Let Q be the charge on the plates.
Energy density = 1/2 x ε0 x E^2
100 = 0.5 x 8.854 x 10^-12 x E^2
E = 4.75 x 10^6 V/m
V = E x d
V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V
C = ε0 A / d
C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F
Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C
Q = 190 nC
Answer:
is the drop in the water temperature.
Explanation:
Given:
- mass of ice,
![m_i=14.7\ g=0.0147\ kg](https://tex.z-dn.net/?f=m_i%3D14.7%5C%20g%3D0.0147%5C%20kg)
- mass of water,
![m_w=324\ g=0.324\ kg](https://tex.z-dn.net/?f=m_w%3D324%5C%20g%3D0.324%5C%20kg)
Assuming the initial temperature of the ice to be 0° C.
<u>Apply the conservation of energy:</u>
- Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.
<u>Now from the heat equation:</u>
![Q_i=Q_w](https://tex.z-dn.net/?f=Q_i%3DQ_w)
......................(1)
where:
latent heat of fusion of ice ![=333.55\ J.g^{-1}](https://tex.z-dn.net/?f=%3D333.55%5C%20J.g%5E%7B-1%7D)
specific heat of water ![=4.186\ J.g^{-1}.^{\circ}C^{-1}](https://tex.z-dn.net/?f=%3D4.186%5C%20J.g%5E%7B-1%7D.%5E%7B%5Ccirc%7DC%5E%7B-1%7D)
change in temperature
Putting values in eq. (1):
![14.7 \times 333.55=324\times 4.186\times \Delta T](https://tex.z-dn.net/?f=14.7%20%5Ctimes%20333.55%3D324%5Ctimes%204.186%5Ctimes%20%5CDelta%20T)
is the drop in the water temperature.
Answer:
So airplane will be 1324.9453 m apart after 2.9 hour
Explanation:
So if we draw the vectors of a 2d graph we see that the difference in angles is = 83 - 44.3 = ![83-44.3=38.7^{\circ}](https://tex.z-dn.net/?f=83-44.3%3D38.7%5E%7B%5Ccirc%7D)
Distance traveled by first plane = 730×2.9 = 2117 m
And distance traveled by second plane = 590×2.9 = 1711 m
We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.
Using the law of cosine,
representing the distance between the planes, we see that:
![d^2=2117^2 + 1711^2 -2\times (2117)\times (1711)cos(38.7)=1755480.2482](https://tex.z-dn.net/?f=d%5E2%3D2117%5E2%20%2B%201711%5E2%20-2%5Ctimes%20%282117%29%5Ctimes%20%281711%29cos%2838.7%29%3D1755480.2482)
d = 1324.9453 m