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jonny [76]
3 years ago
10

Use the equation for motion to answer the question.

Physics
1 answer:
Anika [276]3 years ago
8 0
I choose the option D.
The velocity is constant, so it’s acceleration is 0 m/s^2.
X = 2 + 15 x 1 + 0 = 17 m
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a cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0
Andreas93 [3]

The final velocity is +15.0 m/s

Explanation:

The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

v=u+at

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u is the initial velocity

a is the acceleration

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v=3.0+(1.5)(8.0)=+15.0 m/s

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3 years ago
Determine the components of reaction at the fixed support
denis-greek [22]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are  

    A_{(x)}  = 400  \ N ,  A_{(y)}  = -500  \ N ,  A_{(z)}  = 600  \ N ,  M_x  = 1225 \  N\cdot m , M_y  = 750 \  N\cdot m ,  M_z  = 0 \  N\cdot m

Explanation:

Looking at  the diagram uploaded we see that there are two  forces acting along the x-axis on the fixed support    

   These force are  400 N  and  A_{(x)} [ i.e the reactive force of  400 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(x)}  - 400  = 0

=>     A_{(x)}  = 400  \ N

Looking at  the diagram uploaded we see that there are two  forces acting along the y-axis on the fixed support  

   These force are  500 N  and  A_{(y)} [ i.e the force acting along the same direction with 500 N   ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(y)}  + 500  = 0

=>     A_{(y)}  = -500  \ N

Looking at  the diagram uploaded we see that there are two  forces acting along the z-axis on the fixed support  

       These force are  600 N  and  A_{(z)} [ i.e the reactive force of  600 N  ]

Hence the sum of forces along the x axis is mathematically represented as

        A_{(z)}  - 600  = 0

=>     A_{(z)}  = 600  \ N

Generally taking moment about A along the x-axis we have that

    \sum M_x  = M_x  - 500 (0.75 + 0.5) + 600 ( 1 ) = 0

=>   M_x  = 1225 \  N\cdot m

Generally taking moment about A along the y-axis we have that

    \sum M_y  = M_y  - 400 (0.75 ) + 600 ( 0.75 ) = 0

=>   M_y  = 750 \  N\cdot m

Generally taking moment about A along the z-axis we have that

    \sum M_z  = M_z = 0

=>   M_z  = 0 \  N\cdot m

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