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Rainbow [258]
2 years ago
13

An object moving in the x direction experiences an acceleration of 2.0 m/s2. this means the object (a) travels 2.0 m in every se

cond. (b) is traveling at 2.0 m/s. (c) is decreasing its velocity by 2.0 m/s every second. (d) is increasing its velocity by 2.0 m/s every second.
Physics
2 answers:
Katena32 [7]2 years ago
5 0

(d) is increasing its velocity by 2.0 m/s every second.

An acceleration of 2 m/s² means that the object is is increasing its velocity by 2.0 m/s every second.

<h3>What is an acceleration?</h3>

The rate at which an object's velocity changes over a predetermined period of time is referred to as its acceleration. Acceleration is mathematically defined as an object's change in velocity divided by the amount of time it took for that change to occur.

                                Acceleration = \frac{change in velocity}{time}

The positive sign of the acceleration denotes a change in velocity that is positive and growing, whereas the negative sign denotes a change in velocity that is decreasing.

Consequently, an acceleration of 2 m/s² indicates that the object's velocity is increasing by 2 m/s every second.

Learn more about acceleration here:

brainly.com/question/12550364

#SPJ4

kolbaska11 [484]2 years ago
5 0

option(d) is increasing its velocity by 2.0 m/s every second is the right answer.

An object moving in the x direction experiences an acceleration of           2.0 m/s^{2} this means the object  is increasing its velocity by 2.0 m/s every second.

  • A 2 m/s^{2} acceleration indicates that the item is accelerating at a rate of 2.0 m/s per second.
<h3>What does acceleration mean?</h3>

An object's acceleration is the rate at which its velocity alters over a given amount of time. Calculating acceleration involves dividing the change in velocity of an object by the time it takes for that change to take place.

Acceleration =\frac{change in velocity}{time}

The acceleration's positive sign indicates a change in velocity that is growing and positive, whereas the acceleration's negative sign indicates a change in velocity that is declining.

Consequently, an acceleration of 2 m/s² indicates that the object's velocity is increasing by 2 m/s every second.

To learn more about acceleration visit:

brainly.com/question/12550364

#SPJ4

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An object of mass 2kg is on an incline where there is an applied force of 15N. The
seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is 2.46 m/s^2

Explanation:

a)

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

  • The weight of the object, labelled as mg (where m is the mass of the object and g is the acceleration of gravity), acting downward
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  • The force of friction, F_f, acting down along the plane

b)

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

F_a - F_f - mg sin \theta = ma

where:

F_a = 15 N is the applied force, pushing forward

F_f = 5 N is the frictional force, acting backward

mg sin \theta is the component of the weight parallel to the incline, acting backward, where

m = 2 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

\theta=15^{\circ} is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)

a is the acceleration

Solving for a, we find:

a=\frac{F_a - F_f - mg sin \theta}{m}=\frac{15-5-(2)(9.8)(sin 15^{\circ})}{2}=2.46 m/s^2

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3 years ago
15 POINTS!!!!!
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A fish is 11.9 cm from the front surface of a fish bowl of radius 33 cm. Where does the fish appear to be to someone in air view
antiseptic1488 [7]

Answer:

The fish would appear 42.7 cm on the left side from the front of the bowl.

Explanation:

The fish (object) distance = 11.9 cm, radius of curvature of the bowl = 33 cm. The distance of image of the fish (image distance) can be determined by applying the mirror formula;

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

where f is the focal length of the reflecting surface, u is the object distance and v is the image distance.

But, f = \frac{radius of curvature}{2}

         = \frac{33}{2}

       f = 16.5 cm

Substitute f = 16.5 = \frac{165}{10}, and u = 11.9 = \frac{119}{10} in equation 1;

\frac{10}{165} = \frac{10}{119} + \frac{1}{v}

\frac{1}{v} = \frac{10}{165} - \frac{10}{119}

  = \frac{1190 - 1650}{19635}

\frac{1}{v} = \frac{-460}{19635}

⇒ v  = \frac{19635}{-460}

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    v = 42.7 cm

The fish would appear 42.7 cm on the left side from the front of the bowl.

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3 years ago
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