Answer:
The speed is 
Explanation:
From the question w are told that
The angle made is 
The distance above the surface of the water is 
The value of 
The maximum height attained by the fish is mathematically evaluate as
Making v which is the speed of the fish the subject of the formula
substituting values
![v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }](https://tex.z-dn.net/?f=v%20%3D%20%20%5Csqrt%7B%20%5Cfrac%7B2%2A10%20%2A1.2%20%7D%7B%20%5Bsin%20%2830%29%5D%5E2%20%20%7D%20%7D)
Answer:
1.25 m
Explanation:
From the question given above, the following data were obtained:
Force ratio = 2.5
Distance of load from the fulcrum = 0.5 m
Distance of effort =.?
The distance of the effort from the fulcrum can be obtained as illustrated below:
Force ratio = Distance of effort / Distance of load
2.5 = Distance of effort / 0.5
Cross multiply
Distance of effort = 2.5 × 0.5
Distance of effort = 1.25 m
Therefore, the distance of the effort from the fulcrum is 1.25 m
<u>Answer:</u>
The correct answer option is D. The distance between the planet and the Sun changes as the planet orbits the sun.
<u>Explanation:</u>
Kepler’s laws of planetary motion, derived by the German astronomer Johannes Kepler, are the laws of physics that describe the motions of the planets in the solar system.
According to the Kepler's first law of planetary motion: the path on which the planets orbit around the sun is elliptical in shape, with the center of the sun at one focus.
Therefore, the distance between the Sun and the planets vary as the planet orbit around the sun.
Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:
So now we have an equation and unkown value.
for the thrown rock
for the dropped rock
solving both equation with the quadratic formula:
we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
