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2 years ago

8

you know that there are 1609 meter in a mile. the number of feet in a mile is 5280. what is the speed snail from problem 7 per m

inute.

1 answer:

Zanzabum2 years ago

5 0

Answer:

We know that 1 meter = 100 centimeters, and 1 foot = 12 inches.

So (1,609 meters) x (100 centimeters/meter) = (5,280 feet) x (12 inches/foot)

The second fraction on each side of the equation is equal to ' 1 ', because

the numerator is equal to the denominator, so sticking it in there doesn't

change the value of that side of the equation. But now we can cancel some

units,and wind up with the units we need.

(1,609 meters) x (100 centimeters/meter) = (5,280 feet) x (12 inches/foot)

(1,609 x100) centimeters = (5,280 x 12) inches

160,900 centimeters = 63,360 inches

Divide each side by 63,360 : 2.54 centimeters = 1 inch

Explanation:

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a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

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a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

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$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

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$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

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$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

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$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

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The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

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The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)

Where:

$a$ - Net acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

$\Delta y$ - Covered distance, measured in meters.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $\Delta y = 1.5\,m$ , then the time taken by the system is:

$t = \sqrt{\frac{2\cdot \Delta y}{a} }$

$t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }$

$t \approx 0.845\,s$

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

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$v = a\cdot t$ (4)

Where $v$ is the final speed of the system, measured in meters per second.

If we know that $a = 4.203\,\frac{m}{s^{2}}$ and $t \approx 0.845\,s$ , then the final speed of the system is:

$v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)$

$v = 3.551\,\frac{m}{s}$

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