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adoni [48]
3 years ago
12

Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests o

n a support that serves as an axis of rotation. The block on the right has a mass of 2.8 kg. Determine the magnitude of the angular acceleration when the system is allowed to rotate.
Physics
1 answer:
Andrew [12]3 years ago
7 0

Answer:

The magnitude of the angular acceleration ∝ = \frac{rxF}{2.8[tex]r^{2}}[/tex]

Explanation:

The angular acceleration ∝ is equal to the torque (radius multiplied by force) divided by the mass times the square of the radius. The magnitude of angular acceleration ∝ will have the equation above but we have to replace the mass in the equation by 2.8kg as stated.

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The sales price of a product is $2 per unit; the variable cost is $1 per unit; and fixed costs total $1,000. how many units must
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Answer:

<u><em>1000 units for breakeven</em></u>

Explanation:

Let x be the number of units sold at breakeven.

The total sales at the point would be $2x.

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Total costs are = $1x + $1000

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2 years ago
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
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Answer:

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Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

  • x0 is the initial position.
  • v0 is the initial velocity.
  • a is the acceleration.
  • t is the time interval in which the motion is studied.

In this case:

  • x0= 0
  • v0= 0  because the object is initially stationary
  • a= 6 \frac{m}{s^{2} }
  • t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 \frac{m}{s^{2} }*(15s)²

Solving:

x=½*6 \frac{m}{s^{2} }*(15s)²

x=½*6 \frac{m}{s^{2} }*225 s²

x= 675 m

<u><em> The object will travel 675 m during that time.</em></u>

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