What's the example you are talking about?
Distance and displacement are hardly ever equal.
Remember that 'displacement' is the straight-line distance between
the start-point and the end-point, regardless of what path you followed
on the way.
So they're equal ONLY when the trip from start to finish was completely
in a straight line.
Yes because if energy could have a positive change why can you not have a negative change
a) create an expression
for the ball's initial horizontal velocity, V0x, in terms of the variables
given in the problem statement.
v0x = vf * cos(Θf)
<span>
b) calculate the ball's initial vertical velocity, V0y, in
m/s</span>
v0x = 32.4m/s * cos(-25.5º)
= 29.2 m/s <span>
tanΘ = v1y / v0x → tan(-25.5) = v1y / 29.2m/s → v1y = -13.93
m/s
the vertical velocity when the ball was caught.
(v0y)² = (v1y)² + 2as = (-13.93m/s)² + 2 * 9.8m/s² * 5.5m = 301.78
m²/s²
v0y = 17.37 m/s
c) calculate the magnitude of the ball's initial velocity,
v0, in m/s</span>
v0 = sqrt (v0y^2 +
v0x^2)
v0 = sqrt (17.37^2 + 29.2^2)
m/s
v0 = 33.98 m/s
<span>
d) find the angle, theta0, in degrees above the horizontal at
which which the ball left the bat.</span>
tan Θ = v0y/v0x
<span>Θ = arctan(17.37/29.2) =
30.75º above horizontal</span>
Answer:
the spring constant is -89.2857 n/m
Explanation:
The computation of the spring constant is shown below:
As we know that
Force in newtons = Spring constant × amount of extension
F = -k × x
where
F = 25 N
And, x = 0.280m
So, the spring constant would 2
= 25N ÷ 0.280 m
= -89.2857 n/m
Hence the spring constant is -89.2857 n/m