Answer: Option B. R = (1/2)gt^2
Explanation:
S = R (horizontal distance)
V^2 = 2gS
V^2 = 2gR
R = V^2 / 2g
But V = gt
R = (gt)^2 / 2g
R = (g^2 x t^2) / 2g
R = gt^2 / 2
But t^2 = 2h/g
R = ( g x 2h/g) / 2
R = h
But h = (1/2)gt^2
R = h = (1/2)gt^2
Answer:
If gravity on Earth is increased, this gravitational tugging would have influenced the moon's rotation rate. If it was spinning more than once per orbit, Earth would pull at a slight angle against the moon's direction of rotation, slowing its spin. If the moon was spinning less than once per orbit, Earth would have pulled the other way, speeding its rotation.
Normal force, friction force, gravitational force
Answer:
Explanation:
c. By using the Select Data button and the Select Data Source optionExplanation:A scatter plot is a plot which is used to plot the points of the data on the horizontal and the vertical axis also it depicts how one variable is affected by the another. After preparing the scatter plot to enter the data in the scatter plot we need to use the data button and then data source option so that the data could be entered in the scatter plothence, option c is correct
Answer:
The x-coordinate of the particle is 24 m.
Explanation:
In order to obtain the x-coordinate of the particle, you have to apply the equations for Two Dimension Motion
Xf=Xo+Voxt+0.5axt²(I)
Yf=Yo+Voyt+0.5ayt² (II)
Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.
The particle starts from rest from the origin, therefore:
Vox=Voy=0
Xo=Yo=0
Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:
12=0+(0)t+ 0.5(1.0)t²
12=0.5t²
Dividing by 0.5 and extracting thr squareroot both sides:
t=√12/0.5
t=√24 = 2√6
Replacing t=2√6, ax=2.0,Xo=0 and Vox=0 in (I) to obain the x-coordinate:
Xf=0+0t+0.5(2.0)(2√6)²
Xf= 24 m
