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3 years ago

15

Use differentials to estimate the amount of tin in a closed tin can with diameter 3 inch and height 4 inch, if the top and botto

m are 0.02 inch thick and the side is 0.015 inch thick.

1 answer:

Aloiza [94]3 years ago

3 0

Answer:

dv = 1.03 inch^3

Explanation:

given data:

diameter = 3 inch

radius = 1.5 inch

height 4 inch

top and bottom thickness is 0.02 inch

side thickness = 0.015 inch

we know that volume of the cylinder is given as

$v =\pi r^2 h$

by definition of differential we have

$dv =\frac{\partial v}{\partial r} dr + \frac{\partial v}{\partial h} dh$

where dh = -(0.02 + 0.02) = 0.04 inch [ sum of top and bottom thickness]

the radius is decreased by 0.02 inch, dr = 0.02 inc,

$\frac{\partial v}{\partial r} = 2\pi r h = 37.69$

$\frac{\partial v}{\partial h} = \pi r^2 = 7.06$

dv = 37.69*(0.02) + 7.06*(0.04)

dv = 1.03 inch^3

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3 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

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$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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3 years ago

I started to solve this problem but I’m not sure on what to do next.

snow_tiger [21]

ANSWER:

3408.81 kg

STEP-BY-STEP EXPLANATION:

Given:

v = 111 m/s

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We substitute the values given in the exercise and solve for m (mass)

$\begin{gathered} 21000000=\frac{1}{2}\cdot m\cdot111^2 \\ m=\frac{21000000\cdot2}{111^2} \\ m=3408.81\text{ kg} \end{gathered}$

The mass of the helicopter is 3408.81 kilograms.

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1 year ago