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1 year ago

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Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the met

al in the top and bottom is 0. 1 cm thick and the metal in the sides is 0. 05 cm thick.

1 answer:

ira [324]1 year ago

4 0

The amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal on the top and the bottom is 0.1 cm thick and the metal on the sides is 0.05 cm thick is 8.8 cm.

The formula for calculating the volume of a cylinder is given below.

V = πr^2 h

Get the differential of the volume as shown:

dV = V/ h dh + V / r dr

V/ h = πr^2

V/ h = 2 πr h

Now, the differential becomes

dV = πr^2dh + 2πrh dr

Given the following parameters i.e. diameter and height

dh = 0.1 + 0.1 = 0.2 cm

dr = 0.05 cm

h = 10 cm

d = 4 cm

r = 2cm

Substituting the values in the above equation, we get

dV = 3.14(2)^2(0.2) + 2(3.14)(2)(10)(0.05)

dV = 2.512 + 6.28

dV = 8.792 cm

dV = 8.8 cm

If you need to learn more about diameter click here:

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A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of

Fudgin [204]

Answer:

Part a)

$v_f = v_x = 32.77 m/s$

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

$v_x = 40 cos35 = 32.77 m/s$

$v_y = 40 sin35 = 22.94 m/s$

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

$v_f = 32.77 m/s$

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

$\Delta y = 0$

$0 = v_y t - \frac{1}{2}gt^2$

$T = \frac{2v_y}{g}$

$T = \frac{2(22.94)}{9.81} = 4.68 s$

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3 years ago

Confirm if this is correct or not. If it isn't correct, please correct it.

kow [346]

Answer:

d = 421.83 m

Explanation:

It is given that,

Height, h = 396.9 m

Horizontal speed, v = 46.87 m/s

We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,

$396.9=0\times t+\dfrac{1}{2}\times 9.8 t^2\\\\t=9\ s$

Now d is the distance covered by the cannonball. So,

$d=vt\\\\d=46.87\times 9\\\\d=421.83\ m$

Hence, this is the required solution.

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2 years ago

When is an object magnetic?

Sliva [168]

When it has a magnetic field

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3 years ago

The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t

REY [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is $W = -177.275J$

Explanation:

From the question we are told that

The initial Volume is $Vi = 0.160 L$

The final volume is $V_f = 0.510L$

The external pressure is $P = 5.00 \ atm$

Generally the change in volume is

$\Delta V = V_f - V_i$

Substituting values we have

$\Delta V = 0.510 -0.160$

$= 0.350L$

Generally workdone is mathematically represented as

$W = -P \Delta V$

W is negative because the working is done on the environment by the system which is indicated by volume increase

Substituting values

$W = - 5* 0.350$

$= -1.75 \ L \ \cdot atm$

Now $1 \ L \cdot atm = 101.3J$

Therefore $W = -1.75* 101.3$

$= -177.275J$

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2 years ago

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If we decrease the amount of force applied to an object, and all other factors remain the same, the amount of work completed wil

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A ) decrease.
B ) increase.
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D ) not change.

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Take pushing a box, for example-- You push your hardest then give out, still trying to push the box. You are doing less work than what you have started with!

( Mind marking me for branliest? ; ) )

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3 years ago