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3 years ago

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A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring

shaped solid that remains.

1 answer:

ANTONII [103]3 years ago

5 0

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

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A square-thread power screw is used to raise or lower the basketball board in a gym, the weight of which is W = 100kg. See the f

KIM [24]

Answer:

power = 49.95 W

and it is self locking screw

Explanation:

given data

weight W = 100 kg = 1000 N

diameter d = 20mm

pitch p = 2mm

friction coefficient of steel f = 0.1

Gravity constant is g = 10 N/kg

solution

we know T is

T = w tan(α + φ ) $\frac{dm}{2}$ ...................1

here dm is = do - 0.5 P

dm = 20 - 1

dm = 19 mm

and

tan(α) = $\frac{L}{\pi dm}$ ...............2

here lead L = n × p

so tan(α) = $\frac{2\times 2}{\pi 19}$

α = 3.83°

and

f = 0.1

so tanφ = 0.1

so that φ = 5.71°

and now we will put all value in equation 1 we get

T = 1000 × tan(3.83 + 5.71 ) $\frac{19\times 10^{-3}}{2}$

T = 1.59 Nm

so

power = $\frac{2\pi N \ T }{60}$ .................3

put here value

power = $\frac{2\pi \times 300\times 1.59}{60}$

power = 49.95 W

and

as φ > α

so it is self locking screw

8 0

3 years ago

A race car is travelling around a circular track. The velocity of the car is 20 m/s, the radius of the track is 300 m, and the m

Zielflug [23.3K]

Answer:

μ = 0.136

Explanation:

given,

velocity of the car = 20 m/s

radius of the track = 300 m

mass of the car = 2000 kg

centrifugal force

$F_c = \dfrac{mv^2}{r}$

$F_c = \dfrac{2000\times 20^2}{300}$

F c = 2666. 67 N

F f= μ N

F f = μ m g

2666.67 = μ × 2000 × 9.8

μ = 0.136

so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136

5 0

3 years ago

BTSAUDY5 NAME STARTS FROM THIS

otez555 [7]

Answer:

What's your question?

Explanation:

I'm not able to understand it....

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7 0

3 years ago

Read 2 more answers

A pump with a power of 5 kW (pump power, and not useful pump power) and an efficiency of 72 percent is used to pump water from a

almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water ( $\dot m$ ), in kilograms per second, is determined by following formula:

$\dot m = \frac{\eta \cdot \dot W}{g\cdot H}$ (1)

Where:

$\dot W$ - Pump power, in watts.

$\eta$ - Efficiency, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$H$ - Hydrostatic column, in meters.

If we know that $\eta = 0.72$ , $\dot W = 5000\,W$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the mass flow rate of water is:

$\dot m = 14.683\,\frac{kg}{s}$

The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump ( $\Delta P$ ), in pascals, is determined by this equation:

$\Delta P = \rho\cdot g\cdot H$ (2)

Where $\rho$ is the density of water, in kilograms per cubic meter.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $H = 25\,m$ , then the pressure difference is:

$\Delta P = 245175\,Pa$

The pressure difference across the pump is 245.175 kilopascals.

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The Americans with Disabilities Act requires wheelchair ramps (or other ways for wheelchair users to access a building, such as a wheelchair lift) in new construction for public accommodations in the United States.

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To know more about wheelchair ramps, visit: brainly.com/question/17395685

#SPJ4

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1 year ago