Answer:
Explanation:
From the given information:
Distance 
Speed of the comet 
At distance 
where;
mass of the sun = 

To find the speed
:
Using the formula:

![E_f = E_i + 0 \\ \\ K_f + U_f = K_i + U_i \\ \\ = \dfrac{1}{2}mV_f^2 + \dfrac{-GMm}{d^2} = \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [ \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}](https://tex.z-dn.net/?f=E_f%20%3D%20E_i%20%2B%200%20%5C%5C%20%5C%5C%20%20K_f%20%2B%20U_f%20%3D%20K_i%20%2B%20U_i%20%20%5C%5C%20%5C%5C%20%3D%20%5Cdfrac%7B1%7D%7B2%7DmV_f%5E2%20%2B%20%20%5Cdfrac%7B-GMm%7D%7Bd%5E2%7D%20%3D%20%20%5Cdfrac%7B1%7D%7B2%7DmV_i%5E2%2B%20%5Cdfrac%7B-GMm%7D%7Bd_i%7D%20%5C%5C%20%5C%5C%20V_f%20%3D%20%5Csqrt%7BV_i%5E2%20%2B%202%20GM%20%5CBig%20%5B%20%20%5Cdfrac%7B1%7D%7Bd_2%7D-%20%5Cdfrac%7B1%7D%7Bd_i%7D%5CBig%20%5D%7D)
![V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [ \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}](https://tex.z-dn.net/?f=V_f%20%3D%20%5Csqrt%7B%289.1%20%5Ctimes%2010%5E%7B4%7D%29%5E2%20%2B%202%20%286.67%5Ctimes%2010%5E%7B-11%7D%29%20%2A%281.98%20%2A%2010%5E%7B30%7D%20%29%20%5CBig%20%5B%20%20%5Cdfrac%7B1%7D%7B6%2A10%5E%7B12%7D%7D-%20%5Cdfrac%7B1%7D%7B4.8%2A10%5E%7B10%7D%7D%5CBig%20%5D%7D)

Answer:
B. +m
Explanation:
The magnification of an image is defined as the ratio between the size of the image and of the object:

where we have
y' = size of the image
y = size of the object
There are two possible situations:
- When m is positive, y' has same sign as y: this means that the image image is upright
- When m is negative, y' has opposite sign to y: this means that the image is upside down
Therefore, the correct option representing an upright image is
B. +m
Answer:
i think that it is c because it not only tells us about what the moon does to the earth it also tells us what the earth does to the moon.
Explanation:
Answer:
D = 527.31 Km
Explanation:
given,
angle of ship, θ = 23.5° N of W
distance travel in the direction = 575 Km
Distance of ship in west from harbor = ?
now,
Distance of the ship in the west direction
D = d cos θ
d = 575 Km
θ = 23.5°
inserting all the values
D = 575 x cos 23.5°
D = 575 x 0.91706
D = 527.31 Km
Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km