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Juliette [100K]
3 years ago
9

Which is the best example of translational motion?

Physics
2 answers:
guapka [62]3 years ago
7 0
The best and most correct answer among the choices provided by your question is the second choice or letter B. An example of translational motion is a leaf blowing across a field.

Translational motion<span> is the </span>motion<span> by which a body shifts from one point in space to another. One example of </span>translational motion<span> is the the </span>motion<span> of a bullet fired from a gun. An object has a rectilinear </span>motion <span>when it moves along a straight line.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
sp2606 [1]3 years ago
3 0

Answer:

A leaf blowing across the field.

Explanation:

i took the K12 test! :)

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A Transformer has a primary coil with 650 turns if the output voltage is 500 volts and an input voltage of 200 volts how many te
maxonik [38]

Answer:

1625

Explanation:

Ns/Np = Vs/Vp

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3 years ago
An object moves in one dimensional motion with constant acceleration a = 4.2 m/s^2. At time t = 0 s, the object is at x0 = 3.9 m
solong [7]

Answer:

The object will move to Xfinal = 7.5m

Explanation:

By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:

Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s

With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:

Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =

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8 0
3 years ago
A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15° below the
Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, (\theta) = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ F_x=Fcos(\theta)          and             ⇒ F_y=Fsin (\theta)

⇒ F_x=15cos(15)                           ⇒ F_y=15sin (15)

⇒ Applying concept of  forces.

⇒ \sum F_x=F_n_e_t =F-f

⇒ F_n_e_t =F-f

⇒ ma =F-f       <em>  ...Newtons second law Fnet = ma</em>

⇒ a =\frac{F-f}{m}              

⇒ Plugging the values.

⇒ a =\frac{15cos(15)-0}{40}     <em>...f is the friction which is zero here.</em>

⇒ a =\frac{14.48}{40}

⇒ a=0.362\ ms^-^2

Magnitude of the acceleration of the crate is 0.362 m/s^2.

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