Answer:
It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. ... Once calculated, this area represents the displacement of the object.
Explanation:
Answer:
in the lab
Explanation:
cause that is where scientist spend their time doing research ...
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
Answer:
F=1.65 x 10²⁶ N
Explanation:
Given that
Distance ,R= 3.34 x 10¹² m
Mass m₁= 2.78 x 10³⁰ kg
Mass ,m₂= 9.94 x 10³⁰ kg
we know that gravitational force F given as

G=Constant
G=6.67 x 10⁻¹¹ Nm²/kg²
Now by putting the values

F=1.65 x 10²⁶ N
Therefore the force between these two mass will be 1.65 x 10²⁶ N.