V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
Answer:
45 meters
Explanation:
20 min = 15 meters
So if 20 x 3 = 60
you have to do 3 x 15 !
- which equals to 45 <3
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Explanation:
- Climbing a mountain is similar to hiking from the equator to the pole because in both cases temperature decreases.
- The higher you go, the cooler it becomes.
- For a certain elevation, there is particular drop in temperature. High altitudes offers cooler temperatures.
- The equator receives a huge insolation and the sun is overhead there.
- It implies that the temperature is always high around the equatorial region.
- As one increases latitude, the temperature drops and its is coldest at the pole.
- In both cases, temperature drops and it gets colder.
Learn more:
Temperate and tropics brainly.com/question/10856870
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Answer:
The frequency is 
Explanation:
From the question we are told that
The time taken for it to decay to half its original size is 
Let the voltage of the capacitor when it is fully charged be 
Then the voltage of the capacitor at time t is said to be 
Now this voltage can be mathematical represented as
Where RC is the time constant
substituting values
Generally the cross-over frequency for a low pass filter is mathematically represented as
substituting values
Correct option: A
An object remains at rest until a force acts on it.
As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.
