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Blizzard [7]
3 years ago
7

If the acceleration of a moving object is zero, which of the following best describes its motion

Physics
2 answers:
podryga [215]3 years ago
7 0
If the acceleration of a moving object is zero, the object remains at a constant speed (without friction)
Morgarella [4.7K]3 years ago
3 0

Answer:

Explanation:

According to the newton's second law, the force exerted on an object is equal to the product of the mass of the body and the acceleration of the body.

F = m x a

Where, F is the force and a be the acceleration and m be the mass of the body.

Acceleration is defined as the rate of change of velocity of the body.

As acceleration is zero, so

a = 0

v - u = 0

Where, v is the final velocity of the body and u be the initial velocity of the body.

It means the body is moving with uniform velocity.

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Answer:

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4 0
3 years ago
A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi
Alex_Xolod [135]
  • Magnitude: 12.1 N.
  • Direction: 17.0° to the 8 N force.
<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

  • \displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.
  • \displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.

The sum of forces on each direction will be the resultant force on that direction:

  • Resultant force parallel to the 8 N force: (8 + \dfrac{5\sqrt{2}}{2})\;\text{N}.
  • Resultant force normal to the 8 N force: \dfrac{5\sqrt{2}}{2}\;\text{N}.

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N} (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491.

Find the size of the angle using inverse tangent:

\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree.

In other words, the resultant force is 17.0° relative to the 8 N force.

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