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2 years ago

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An engineer wants to double the water supply reserve and makes a tank 1.26 times as big as the old one in linear dimension, and

the diameter of the legs is 1.41 times as large. Is this big enough?

1 answer:

Fiesta28 [93]2 years ago

8 0

Answer:

Yes, it would be enough.

Explanation:

The volume of the tank (cylinder tank) is given by the following equation:

$V_{initial}=\pi (\frac{D}{2})^{2}h$

Where:

D is the diameter of the tank

h is the height of the tank

Now, the engineer makes a tank 1.26 times as big in heigh and 1.41 times as large with respect to diameter, then the new volume of this tank will be:

$V_{new}=\pi (\frac{1.41D}{2})^{2}1.26h$

$V_{new}=1.41^2*1.26\pi (\frac{D}{2})^{2}h$

$V_{new}=2.51\pi (\frac{D}{2})^{2}h$

in terms of the initial volume:

$V_{new}=2.51V_{initial}$

Which means that the volume of the water could be doubled.

I hope it helps you!

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Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first d

MrMuchimi

Answer:

The width of the slit is 0.4 mm (0.00040 m).

Explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 × $10^{-9}$ m)

d = (3.30 × 563 × $10^{-9}$ ) ÷ (0.0047)

= 1.8579 × $10^{-6}$ ÷ 0.0047

= 0.0003951 m

d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).

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3 years ago

Maceo is making rock candy. Which best describes the steps she should take?

Andru [333]

Answer:

The answer is heat a saturated sugar water solution, dissolve more sugar, then let the solution cool

Explanation:

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3 years ago

Read 2 more answers

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the

german

Answer: 4 s

Explanation:

Given

The ball leaves the hand of student with a speed of $u=19\ m/s$

When the hand is $h=2.5\ m$ above the ground

Using the equation of motion we can write

$h=ut+\dfrac{1}{2}at^2$

Substitute the values

$\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]$

Thus, the ball will take 4 s to hit the ground.

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2 years ago

g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr

Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

$a = \frac{GM}{r^2}$

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

$\frac{da}{dr} = -2\frac{GM}{r^3}$

$da = -2\frac{GM}{r^3}dr$

At the same time since Newton's second law we know that:

$F_w = ma$

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

$F_W = m (\frac{GM}{r^2} ) = 600N$

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

$dF_W = mda$

$dF_W = m(-2\frac{GM}{r^3}dr)$

$dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})$

$dF_W = -2F_W(\frac{dr}{r})$

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

$dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})$

$dF_W = -0.3N$

Therefore there is a weight loss of 0.3N every kilometer.

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3 years ago

Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center o

jolli1 [7]

Answer:

c)At a distance greater than r

Explanation:

For a satellite in orbit around the Earth, the gravitational force provides the centripetal force that keeps the satellite in motion:

$G\frac{Mm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance between the satellite and the Earth's centre

v is the speed of the satellite

Re-arranging the equation, we write

$r = \frac{GM}{v^2}$

so we see from the equation that when the speed is higher, the distance from the Earth's centre is smaller, and when the speed is lower, the distance from the Earth's centre is larger.

Here, the second satellite orbit the Earth at a speed less than v: this means that its orbit will have a larger radius than the first satellite, so the correct answer is

c)At a distance greater than r

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3 years ago