Answer:
2.91 x 10¹² sec
Explanation:
d = distance of nearest star, Proxima Centauri = 4.3 ly = 4.3 x 9.46 x 10¹⁵ m
v = speed of new horizon probe = 14 km/hr = 14000 m/s
t = time taken for the new horizon probe to reach nearest star, Proxima Centauri = ?
Using the equation
d = v t
Inserting the values given
4.3 x 9.46 x 10¹⁵ = (14000) t
t = 2.91 x 10¹² sec
Well,
When an object's velocity changes, we call it acceleration.
Acceleration: The time rate of change in an object's velocity
Since both hv same mass and elsstic collision, so their velocity will exchange. Bob A will stop and bob B will move with speed of A just before the collision.
Speed will be = squreroot ( 2*g*L)
L is length of pendulum
Answer:
a = 1.152s
b = 0.817 m
c = 7.29m/s
Explanation: let the following
From the first equation of linear motion
V = u+at..........1
parameters be represented as :
t = Time taken
v = Final velocity
a = Acceleration due to gravity = 9.8m/s²
u = Initial velocity = 4 m/s
s = Displacement
V = 0
Substitute the values into equation 1
0 = 4-9.8(t)
-4 = -9.8t
t = 4/9.8
t = 0.408s
From : s = ut+1/2at^2.........2
S = 4×0.408+0.5(-9.8)×0.408^2
S= 1.632-4.9(0.166)
S = 1.632-0.815
S = 0.817m
Her highest height above the board is 0.817 m
Total height she would fall is 0.817+1.90 = 2.717 m
From equation 2
s = ut+1/2at^2
2.717 m = 0t+0.5(9.8)t^2
2.717 m = 0+4.9t^2
2.717 m = 4.9t^2
2.717/4.9 = t^2
0.554 =t^2
t =√0.554
t = 0.744s
Hence, her feet were in the air for 0.744+0.408seconds
= 1.152s
Also recall from equation 1
V= u+at
V = 0+9.8(0.744)
V = 7.29m/s
Hence, the velocity when she hits the water is 7.29m/s
Finally,
a = 1.152s
b = 0.817 m
c = 7.29m/s
The instant it was dropped, the ball had zero speed.
After falling for 1 second, its speed was 9.8 m/s straight down (gravity).
Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.
Falling for 1 second at an average speed of 4.9 m/s, is covered <em>4.9 meters</em>.
ANYTHING you drop does that, if air resistance doesn't hold it back.
