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kodGreya [7K]
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.
Physics
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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A train moves from rest to a speed at 25m/s in 30.0 seconds. What is it’s acceleration?
sasho [114]

Answer:

25/30 = 5/6 m/s^2 5/6 meters per second squared

4 0
3 years ago
Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi
Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}

here we know that

\mu_1 = 1.473

h_o = 5 cm

\mu_2 = 1

now by above formula

\frac{1.473}{5} = \frac{1}{h_i}

h_i = 3.39 cm

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0
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

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\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.

5 0
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

Ek = 21000000 J

We have that the formula for kinetic energy is as follows:

E_k=\frac{1}{2}\cdot m\cdot v^2

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.

7 0
1 year ago
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