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Vadim26 [7]
3 years ago
12

Would the springs inside a bathroom scale be more compressed or less compressed if you weighed yourself in an elevator that acce

lerated upward? Accelerated downward? Would the springs inside a bathroom scale be more compressed or less compressed if you weighed yourself in an elevator that accelerated upward? Accelerated downward? upward, more compressed; downward, more compressed upward, less compressed; downward, less compressed upward, more compressed; downward, less compressed upward, less compressed; downward, more compressed
Physics
1 answer:
Delvig [45]3 years ago
5 0

Answer:

Accelerate up more compressed spring

Accelerate down spring less compressed

Explanation:

For this problem we must analyze the forces that act on the scale on the one hand the weight of the man directed downwards and I have the other the elastic force of the spring directed upwards. Now let's write Newton's second law for these forces in various configurations,

When the elevator is quiet or moving at constant speed

       Fe - W = 0      ⇒    Fe = W         -k x = mg    x = mg /k

       Fe = mg

We use this value to compare

Now let's analyze when the elevator accelerates upwards

      Fe -W = m a

     Fe = ma + W

     Fe = m (a + g)

So we can see that Fe increases, so the compression of springs is higher

Now let's analyze when the elevator acceleration is down

    Fe -W = m (-a)

    Fe = w - m a

    Fe = m (g -a)

In this case Fe is smaller, so the compression of the spring is less

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Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh
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Answer:

V₀ = 5.47 m/s

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3 years ago
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
svetoff [14.1K]

Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet x=\frac{4}{12} =\frac{1}{3}feet

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

W=kx ⇒ k=\frac{W}{x}

The spring constant , k=\frac{24}{1/3}

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    m\frac{d^2x}{dt} +kx=0

    \frac{3}{4} \frac{d^2x}{dt} +72x=0

  \frac{d^2x}{dt} +96x=0

Auxiliary equation is, m^2+96=0

                                 m=\sqrt{-96}

                               =\frac{+}{} i4\sqrt{6}

Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

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                    =-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2 4\sqrt{6} cos4\sqrt{6(0)} =0

                                                    c_2 4\sqrt{6} =0

                                     c_2=0

Therefore , x(t)=c_1 cos 4\sqrt{6t}

Since , the mass is released from the rest from 4 inches

                    x(0)= -4 inches

c_1 cos 4\sqrt{6(0)} =-\frac{4}{12} feet

   c_1=-\frac{1}{3} feet

Therefore , the equation of motion is  -\frac{1}{3} cos4\sqrt{6t}

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