I think it is c I'm only in 7th grade but I'm pretty sure that the answer is c
Answer
given,
mass of the ball = 3 kg
swing in vertical circle with radius = 2 m
work done by the gravity = ?
work done by the tension = ?
Work done by the gravity = - m g Δh
Δ h = 2 + 2 = 4 m
Work done by the gravity =
= -117.6 J
work done by gravity is equal to -117.6 J
Work done by tension will be equal to zero.
Zero because tension is always perpendicular to velocity
work done by tension is equal to 0 J
<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:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
