Answer:
x coordinate = -1.66 m
y coordinate is = -0.825m
Explanation:
Suppose z be the distance form the first charge and z + sqrt(1^2 +.5^2) be the distance from the second So z + sqrt(1+.25) = z + 1.12
We have k*2.0x10^-6/s^2 = k*6x10^-6/(s+1.12)^2
0.0356s^2 -0.019s-0.0897=0
s=1.876m
The angle of the line between the two charges is arctan(.5/1) = 26.6o
x coordinate = -1.876*cos(26.6) = -1.66m
y coordinate is -1.876*sin(26.6) = -0.825m
The planet closest to the sun; Mercury.
Answer:
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Explanation:
Hi there!
The equations of height and velocity of the ball are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity of the ball at time t.
Placing the origin at the throwing point, y0 = 0.
Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.
v = v0 + g · t
6.00 m/s = 12.0 m/s -9.81 m/s² · t
(6.00 - 12.0)m/s / -9.81 m/s² = t
t = 0.612 s
Now, let´s calculate the height of the baseball at that time:
y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)
y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²
y = 5.51 m
The ball will have an upward velocity of 6 m/s at a height of 5.51 m.
Have a nice day!
Answer:
I believe D
Explanation:
You need to have a more accurate reading and you want to test it multiple times throughout the week though to get a base resting rate.
I hope this is correct good luck!
