Answer:
1992 (Early 1990s)
Explanation:
First of all, i would like to define an extrasolar planet as a planet that orbits a start that is not our own.
The first confirmed detections of extrasolar planets occured in the early 1990s (specifically 1992, some say 1995). The name of the first extrasolar planet is widely believed to be called Dimidium or 51 Pegasi b.
Extrasolar were searched by monitoring stars for slight dimming that might occur as unseen planets pass in front of them.
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
<h2>
Answer:50</h2>
Explanation:
Let be the airspeed.
Let be the cross wind speed.
We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.
If and are two perpendicular vectors,the resultant vector has the magnitude
Given,
So,the ground speed is