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2 years ago

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A sheet of aluminum (al) foil has a total area of 1.000 ft2 and a mass of 3.716 g. what is the thickness of the foil in millimet

ers (density of al = 2.699 g/cm3)?

1 answer:

Andrews [41]2 years ago

3 0

The mass of the sheet of aluminium is the product between its density d and its volume V:
$m=dV$
Re-arranging the equation, we can find the volume of the sheet:
$V= \frac{m}{d}= \frac{3.716 g}{2.699 g/cm^3}= 1.377 cm^3$

Converted in millimeters,
$V=1377 mm^3$

The area is
$A=1.00 ft^2$
Since $1 ft = 0.305 m$ , this area corresponds to
$A=1.00 ft^2 = (0.305 m)^2 = 0.093 m^2 = 0.093 \cdot 10^6 mm^2$

So we can now find the thickness of the sheet of aluminium:
$t= \frac{V}{A}= \frac{1377 mm^2}{0.093 \cdot 10^6 mm^2}=0.015 mm$

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1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

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In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

3)

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

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3 years ago

Maurice pulls on the end of a spring scale. He lets go of the end and observes the spring snap back into place. What force resto

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Elastic is the right answer.

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2 years ago

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Answer:

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where,

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g = acceleration due to gravity = 9.8 m/s²

Therefore,

Tension = (0.00127 kg)(9.8 m/s²)

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Sarah's acceleration is $-0.49 m/s^2$

Explanation:

The force of kinetic friction acting on Sarah has a magnitude which is given by:

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where

$\mu$ is the coefficient of kinetic friction

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Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

$F=ma$

where a is the acceleration

Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:

$F=-\mu mg = ma$

where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find

$a=-\mu g$

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we find:

$a=-(0.05)(9.81)=-0.49 m/s^2$

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2. Inelastic collision: When the momentum the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

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