The answer is he weighs 187.39 LBS/Pounds
<h2>
Answer: The Transit method</h2>
Detecting extrasolar planets by direct observation (with a telescope) is a complicated task. This is because any planet constitutes an extremely dim light source compared to the star around which it orbits.
So, to detect this extremely dim source is quite difficult due to the glare of the star's light that dulls it.
In this sense, scientists and astronomers have made several methods to find these extrasolar planets, among which the most successful has been the transit method.
This method is based on <u>astronomical transit</u>, a phenomenon in which a body (a planet in this case) passes in front of a larger one (the star), blocking (eclipsing) its vision to some extent.
It should be noted that this is the method currently used in the search for extrasolar planets. Space agencies such as ESA (Europe) and NASA (USA) have put into orbit satellites with extremely sensitive photometric sensors to observe even the smallest variations of intensity of a star due to the passage of a planet.
Answer:
The rock's speed after 5 seconds is 98 m/s.
Explanation:
A rock is dropped off a cliff.
It had an initial velocity of 0 m/s. And now it is moving downwards under the influence of gravitational force with the gravitational acceleration of 9.8 m/s².
Speed after 5 seconds = V
We know that acceleration = average speed/time
In our case,
g = ((0+V)/2)/5
9.8*5 = V/2
=> V = 2*9.8*5
V = 98 m/s
<span>D. Pressure increases with increasing depth.
This occurs because there is more weight above you to increase the pressure.
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Answer:
The velocity of the ball before it hits the ground is 381.2 m/s
Explanation:
Given;
time taken to reach the ground, t = 38.9 s
The height of fall is given by;
h = ¹/₂gt²
h = ¹/₂(9.8)(38.9)²
h = 7414.73 m
The velocity of the ball before it hits the ground is given as;
v² = u² + 2gh
where;
u is the initial velocity of the on the root = 0
v is the final velocity of the ball before it hits the ground
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 7414.73 )
v = 381.2 m/s
Therefore, the velocity of the ball before it hits the ground is 381.2 m/s
