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alina1380 [7]
3 years ago
5

A cylindrical metal can is to have no lid. It is to have a volume of 125π in3. What height minimizes the amount of metal used?

Physics
1 answer:
Alex Ar [27]3 years ago
6 0

Answer:

Minimum height of metal = 5 inches

Explanation:

Volume of the cylindrical metal = πR²H = 125π

cancelling out π on both sides

R²H = 125

Hence it can be deduced that R² = 25 and H = 5

Hence minimum height of metal = 5 inches

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The higher density allows the protogalactic clouds to cool faster and form an elliptical galaxy rather than a spiral galaxy.

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Composition

Since there had been no previous star formation to create other elements, protogalaxies would have been made up almost entirely of hydrogen and helium. The hydrogen would bond to form H2 molecules, with some exceptions. This would change as star formation began and produced more elements through the process of nuclear fusion.

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Once a protogalaxy begins to form, all particles bound by its gravity begin to free fall towards it. The time taken for this free-fall to conclude can be approximated using the free-fall equations. Most galaxies have completed this free-fall stage to become stable elliptical or disk galaxies, the disks taking longer to fully form. The formation of galaxy clusters takes much longer and is still in progress now.

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Learn more about protogalactic clouds here : brainly.com/question/28166070

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7 0
2 years ago
An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode
sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell v_{0} in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

                v_{x}=v_{0} \times \cos \theta

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of v_{x} and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

           v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}

           v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}

For motion with constant acceleration, we know

            s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}

Along the horizontal, x-axis, we might write this as

            x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}

Measuring distances relative to the firing point means

               x_{0}=0

we know that,

              a_{x}=0

or,

             v_{x}=v_{x 0}=\text { constant }

By applying the values, we get,

           x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}

The acceleration of gravity is vertically downward and is g=-9.8 \mathrm{m} / \mathrm{s}^{2} , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

           y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}

we know, y_{0}=0 and a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}, so,

          y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)

                 y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0
3 years ago
Hypsometer (upper fixed point​
aivan3 [116]

Answer:a device for calibrating thermometers at the boiling point of water at a known height above sea level or for estimating height above sea level by finding the temperature at which water boils.

Explanation:

3 0
3 years ago
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ahrayia [7]

Answer:

The object will rotate with constant angular acceleration

Explanation:

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∑iτi=Iα.

.Therefore when constant net torque is applied to object that is rotating, the object will rotate with constant angular acceleration

7 0
3 years ago
Your bike starts with an initial velocity of 1 m/s and you accelerate to 4 m/s in 6 seconds. How far did you travel while accele
garri49 [273]

Answer:

78 m

Explanation:

Initial v = 1 m/s

a = 4 m/s²

t = 6s

x = v_{i} t + \frac{1}{2}at^{2} \\x = (1)(6) + \frac{1}{2} (4)(6)^{2} \\x = 6 + 72\\x = 78

5 0
3 years ago
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