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3 years ago

A- 1000 m/s2 Xi-0m Xf-0.75m Vf-?

Physics

1 answer:

sleet_krkn [62]3 years ago

8 0

Answer:

The final velocity of the object is, $v_{f}$ = 27 m/s

Explanation:

Given,

The acceleration of the object, a = 1000 m/s²

The initial displacement of the object, $x_{i}$ = 0 m

The final displacement of the object, $x_{f}$ = 0.75 m

The initial velocity of the object will be, $v_{i}$ = o m/s

The final velocity of the object, $v_{f}$ = ?

The average velocity of the object,

v = ( $x_{f}$ - $x_{i}$ )/ t

= 0.75 / t

The acceleration is given by the relation

a = v / t

1000 m/s² = 0.75 / t²

t² = 7.5 x 10⁻⁴

t = 0.027 s

Using the I equation of motion,

$v_{f}$ = u + at

Substituting the values

$v_{f}$ = 0 + 1000 x 0.027

= 27 m/s

Hence, the final velocity of the object is, $v_{f}$ = 27 m/s

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QUESTION 10

Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

$\theta =37.01^{\circ}$

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

$T_{1}sin(\theta)-T_{2}sin(\theta)=0$ (1)

Where:

T(1) is the tension due to the rope 1
T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

$T_{1}cos(\theta)+T_{2}cos(\theta)-W=0$ (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

$T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)=m_{steel}g$

T(1) must be equal to 5479 N, so we have:

$cos(\theta)=\frac{m_{steel}g}{2T_{1}}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=0.80$

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

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2. Har (every)

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3 years ago

A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around

Alex_Xolod [135]

Answer:

$\alpha =\frac{m*g*R}{I-m*R^2}$

$a = \frac{m*g*R^2}{I-m*R^2}$

$T=\frac{I*m*g}{I-m*R^2}$

Explanation:

By analyzing the torque on the wheel we get:

$T*R=I*\alpha$ Solving for T: $T=I/R*\alpha$

On the object:

$T-m*g = -m*a$ Replacing our previous value for T:

$I/R*\alpha-m*g = -m*a$

The relation between angular and linear acceleration is:

$a=\alpha*R$

So,

$I/R*\alpha-m*g = -m*\alpha*R$

Solving for α:

$\alpha =\frac{R*m*g}{I+m*R^2}$

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$a =\frac{R^2*m*g}{I+m*R^2}$

And finally, the tension will be:

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These are the values of all the variables: α, a, T

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