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Elis [28]
1 year ago
13

An object, which is at the origin at time t=0

Physics
2 answers:
Paladinen [302]1 year ago
7 0

At time t=0, an object at the origin has a constant acceleration a=(6.0i + 3.0j)m/s2 and an initial velocity V0= (-14.0i - 7.0j)m/s (-16.3 I - 8.2 j) The position x at which the object comes to rest is denoted by m. (temporarily).

<h3>What exactly is Origin?</h3>

A body's origin of motion is defined as the point at which a body or object begins to move.

  • The act of being born, existing, or beginning; point of departure.
  • That causes something to happen.
  • That which serves as a source or is derived from something.
  • In the field of coordinate geometry, the origin is defined as the initial point or starting point from which we begin our calculations or measurements. We start at zero on a ruler.
  • Because the 0 on a ruler is where we start measuring, it is referred to as the scale's origin.
  • The center of a coordinate axis is defined as coordinate 0 in all axes.

Hence, at time t=0, an object at the origin has an initial velocity V0 of (-14.0i - 7.0j)m/s and a constant acceleration an of (6.0i + 3.0j)m/s2. r = (-16.3 I - 8.2 j) m is the position x at which the object comes to rest (temporarily).

  • To learn more about
  • origin
  • refer to:
  • brainly.com/question/28515326
  • #SPJ9
zheka24 [161]1 year ago
6 0

An object, which is at the origin at time t=0, has initial velocity V₀= (-14.0i - 7.0j)m/s and constant acceleration a=(6.0i + 3.0j)m/s². The position x where the object comes to rest (momentarily) is, r = ( -16.3 i - 8.2 j ) m

<h3>What is position?</h3>

Position is a term that means a object where placed. In other words a object is placed or stayed or comes to rest that place called the position of the object.

<h3 /><h3>How can we calculate the position?</h3>

To calculate the position of the object we are using two formulas, like

V= V₀+∫a dt

r=r₀+∫v dt

Here we are given,

V₀=initial velocity of the object=(-14.0i - 7.0j)m/s,

a=constant acceleration of the object= (6.0i + 3.0j)m/s²

Now we put the values in the first formula,

V= V₀+∫a dt

Or, V= (-14.0i - 7.0j)+ ∫(6.0i + 3.0j) dt

Or, V= (-14.0+6.0t)i + (- 7.0+ 3.0t)j

According to the question, the particle becomes rest then,

V(x)=0

Or, (-14.0+6.0t)=0

Or, t=7/3S

In other way, V(y)=0

Or, (- 7.0+ 3.0t)=0

Or, t=7/3S

Now we put the value of velocity in the second equation,

r=r₀+∫v dt

Or, r=0+∫(-14.0+6.0t)i + (- 7.0+ 3.0t)j dt

Or, r= (-14.0t+6.0t²)i + (- 7.0t+ 3.0t²)j

Now we put,  t=7/3S, then the r value becomes,

r= (-14.0(7/3)+6.0(7/3)²)i + (- 7.0(7/3)+ 3.0(7/3)²)j

r≈(-16.3 i - 8.2 j)m

From the above calculation we can conclude that, The position x where the object comes to rest (momentarily) is, r = ( -16.3 i - 8.2 j ) m

Learn more about  position:

brainly.com/question/2534565

#SPJ9

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A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16
erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle \phi = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

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F = \frac{1200}{2.93}

F = 408.73 N

Now the bending stress is given by the formula

\sigma = \frac{K_{v} F}{m b y}

\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}

\sigma = 35.38 M pa

This is the value of bending stress on the pinion

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