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3 years ago

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could you help me with this question: Aspherical shell is formed by taking a solid sphere of radius 24.0 cm and hollowing out a

spherical section from the shell’s interior. Assume the hollow section and the sphere itself have the same center location (that is, they are concentric). a) If the hollow section takes up 86.0 percent of the total volume, what is its radius? b) What is the ratio of the outer area to the inner area of the shell?

1 answer:

kumpel [21]3 years ago

5 0

Answer:

La vdd tengo la misma duda, de q prepa eres?

Explanation:

Nada más para saber

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For this exercise, use the position function s(t) = −4.9t2 + 250, which gives the height (in meters) of an object that has falle

nignag [31]

Answer:

t=7.14s

v=-69.972 m/s

Explanation:

Position function

$s(t)=-4.9*t^{2}+250$

Velocity is the derivative of position function

$V(t)=\frac{dx}{dt}\\V(t)=-2*4.9*t\\V(t)=-9.8*t$

The time the object hit the ground can be find by the given function know that the position is going to be 0m

$s(t)=-4.9*t^{2}+250$

$s(t)=0\\0=-4.9*t^{2} +250\\t=\sqrt{\frac{250}{4.9}}\\t=7.14s$

Check:

$s(7.14)=-4.9*(7.14s)^{2}+250\\ s(7.14)=-250+250\\s(7.14)=0m$

So the velocity can be find using the time discovery before and using the same function but with the derivate

$V(t)=-2*4.9*t\\V(7.14)=-2*4.9*(7.14)\\V(7.14)=-69.972 \frac{m}{s}$

The velocity is negative because the object is moving downward

6 0

3 years ago

A cable that weighs 4 lb/ft is used to lift 1000 lb of coal up a mine shaft 700 ft deep. Find the work done.

Shkiper50 [21]

Answer:

<h2>980000ft-lbs</h2>

Explanation:

Step one:

given data

mass of cable= 4lb/ft

mass of coal= 1000lb

dept of mine= 700ft

Step two:

Required

the work-done to lift the coal and the rope combined

Work-done to lift coal

Wc=1000*700= 700,000 lb-ft

Work-done to lift rope

$Wr=\int\limits^{700} _0 {4(700-y)} \, dx \\\\Wr=4(700y-\frac{1}{2}y^2 )\limits^{700}_0$

substitute y=700 we have, since y=0 will result to 0

$Wr=4(700*700-\frac{1}{2}*700^2 )\\\\Wr=4(490000-245000)\\\\Wr=4(245000)\\\\Wr=980000ft-lbs$

6 0

3 years ago

A train travels from over city to another its initial velocity is lower than its final velocity this is a example of

Anni [7]

I belive the answer is D)

4 0

3 years ago

(33%) Problem 3: Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1100 kg and i

expeople1 [14]

Answer:

The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

Explanation:

Given that,

Mass of first car = 1100 kg

Velocity of first car = 8.5 m/s

Mass of second car = 650 kg

Velocity of second car = 17.5 m/s

Suppose we need to find the final velocity of the cars and direction of the cars.

We need to calculate the velocity of the car in west direction

Using conservation of momentum in west direction

$m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{x}$

$v_{x}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}$

Put the value into the formula

$v_{x}=\dfrac{1100\times0+650\times17.5}{1100+650}$

$v_{x}=6.5\ m/s$

We need to calculate the velocity of the car in south direction

Using conservation of momentum in south direction

$m_{f}v_{f}+m_{s}v_{s}= (m_{f}+m_{s})v_{y}$

$v_{y}=\dfrac{m_{f}v_{f}+m_{s}v_{s}}{(m_{f}+m_{s})}$

Put the value into the formula

$v_{y}=\dfrac{1100\times8.5+650\times0}{1100+650}$

$v_{y}=5.3\ m/s$

We need to calculate the final velocity of the cars

Using formula of velocity

$v_{eq}=\sqrt{(6.5)^2+(5.3)^2}$

$v_{eq}=8.38\ m/s$

We need to calculate the direction

Using formula of direction

$\tan\theta=\dfrac{v_{y}}{v_{x}}$

Put the value into the formula

$\tan\theta=\dfrac{5.3}{6.5}$

$\theta=\tan^{-1}(\dfrac{5.3}{6.5})$

$\theta=39.1^{\circ}$

Hence, The final velocity of the cars is 8.38 m/s at an angle 39.1° south of west.

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3 years ago

Distance v. Time

Allushta [10]

Answer:

4

(m)

2 ( s )

Explanation:

ok...........

3 0

4 years ago