An object with mass 110 kg moved in outer space. When it was at location < 11, -35, -7 > its speed was 17.5 m/s. A single constant force < 230, 420, -230 > N acted on the object while the object moved to location < 16, -41, -13 > m. What is the speed of the object at this final location?

final speed =

To determine the final speed of the object at the given final location, we need to calculate the change in velocity and then find the magnitude of the final velocity vector.

The change in velocity can be calculated using the formula:
Δv = F * t / m

where:
Δv is the change in velocity,
F is the force acting on the object,
t is the time for which the force was applied,
and m is the mass of the object.

Given:
Force, F = <230, 420, -230> N
Mass, m = 110 kg

The time for which the force was applied is not provided, so assuming it is the same throughout the motion, we can calculate it using the following formula:
Δx = v * t

where:
Δx is the displacement,
v is the initial speed,
and t is the time.

Given:
Initial velocity, v = 17.5 m/s
Initial position, r1 = <11, -35, -7>
Final position, r2 = <16, -41, -13>

Using the distance formula:
Δx = |r2 - r1| = √[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]

Calculating the displacement:
Δx = √[(16 - 11)^2 + (-41 - (-35))^2 + (-13 - (-7))^2]
= √[5^2 + (-6)^2 + (-6)^2]
= √[25 + 36 + 36]
= √97

Now, let's calculate the time:
Δx = v * t
√97 = 17.5 * t
t = √97 / 17.5

Calculating the change in velocity:
Δv = F * t / m
Δv = <230, 420, -230> * (√97 / 17.5) / 110
Δv = <230 * (√97 / 17.5), 420 * (√97 / 17.5), -230 * (√97 / 17.5)>
Δv ≈ <39.74, 72.39, -39.74>

Finally, let's find the magnitude of the final velocity vector:
|v2| = |v1 + Δv| = √[(v1x + Δvx)^2 + (v1y + Δvy)^2 + (v1z + Δvz)^2]

Given:
Initial velocity, v1 = 17.5 m/s
Change in velocity, Δv = <39.74, 72.39, -39.74>

Calculating the final speed:
|v2| = √[(17.5 + 39.74)^2 + (0 + 72.39)^2 + (0 + -39.74)^2]
|v2| = √[57.24^2 + 72.39^2 + (-39.74)^2]
|v2| ≈ √[3276.8976 + 5243.1121 + 1580.1476]
|v2| ≈ √10100.1573
|v2| ≈ 100.5 m/s

Therefore, the speed of the object at the final location is approximately 100.5 m/s.

To find the final speed of the object at the given final location, we can use the concept of work and energy.

First, we need to calculate the work done by the constant force on the object. The work is given by the dot product of the force and the displacement:
Work = Force * Displacement * cosine(theta)

Given Force = <230, 420, -230> N and Displacement = <16, -41, -13> m, we can calculate the work done:
Work = (230 * 16) + (420 * -41) + (-230 * -13)

Next, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Since the object has no initial kinetic energy (as it is at rest), the work done by the force will be equal to the final kinetic energy.

Final kinetic energy = Work

Now, we can relate the final kinetic energy to the final speed of the object using the formula:
Final kinetic energy = (1/2) * mass * final speed^2

Rearranging the equation, we can solve for the final speed:
final speed = sqrt((2 * Work) / mass)

Given mass = 110 kg and Work (calculated in the previous step), we can substitute the values into the equation to find the final speed of the object at the final location.

The force was applied in the direction

<16,-41,-13> - <11,-35,-7>
F=ma gives you the magnitude of the acceleration

Then just figure t to cover the given distance, and

v = v0 + at to get the final velocity.