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ikadub [295]
1 year ago
5

a lunar probe is moving 1670 m/s at a 73.0 angle. it needs to land on the moon 3.71 x 10^8 m away in a 46.0 direction in 8.64 x

10^4 s. what is the magnitude of the acceleration that the engine must produce?

Physics
1 answer:
Oduvanchick [21]1 year ago
8 0

The magnitude of the acceleration that the engine must produce is  0.065 m/s².

<h3>What is acceleration?</h3>

Acceleration is given by the ratio of Resultant or total force acting on any object and the its mass.

It can also be defined as the rate change of velocity with time.

acceleration a = (Δv) / (Δt)

Given is a lunar probe is moving 1670 m/s at a 73° angle. it needs to land on the moon  3.71 x 10⁸ m away in a 46° direction in 8.64 x 10⁴ s.

The initial velocity will be

1670 x cos (73°-46°) = 1487.981 m/s

Using the second equation of motion, we have

s = ut + 1/2 at²

The final velocity will be zero after landing on Moon.

Substituting the values, we have

3.71 x 10⁸=  1487.981 x  8.64 x 10⁴ + 1/2 a ( 8.64 x 10⁴)²

Solving the equation, we get

acceleration, a = 0.065 m/s²

Thus, the magnitude of the acceleration that the engine must produce is   0.065 m/s²

Learn more about acceleration.

brainly.com/question/12550364

#SPJ1

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Explanation:

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It is also defined as the cross product of force vector and the displacement vector.

\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}

\overrightarrow{\tau }=(- 7 x i + yj)\times (-mgj)

[tex]\overrightarrow{\tau  }= 7 m g x k

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A 782.10 kg car is brought from 7.60 m/s to 3.61 m/s over a time period of 4.23 seconds. What is the average force the car exper
alexandr402 [8]

Answer:

–735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 782.10 kg

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Force (F) =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Acceleration (a) =?

a = (v – u) / t

a = (3.61 – 7.60) / 4.23

a = –3.99 / 4.23

a = –0.94 m/s²

Finally, we shall determine the force experienced by the car as shown below:

Mass (m) of car = 782.10 kg

Acceleration (a) = –0.94 m/s²

Force (F) =?

F = ma

F = 782.10 × –0.94

F = –735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

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Answer:

1. the force which can be felt or act only when two objects are in contact is known as contact force.

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2. the force which can be felt or act even when two objects are in contact or not is known as non-contact force.

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