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adoni [48]
1 year ago
15

Find the cost of excavating a space 84 ft long, 42 ft wide, and 9 ft deep at a cost of $39/yd3. (simplify your answer completely

.)
Physics
1 answer:
m_a_m_a [10]1 year ago
4 0

The cost of excavating a space of 84 ft long, 42 ft wide, and 9 ft deep is $45864

Information about the problem:

  • Space long= 84 ft
  • Space wide= 42 ft
  • Space deep= 9 ft
  • Cost by yard3 = $39/yd3
  • Total cost= ?

To solve this problem, we have to state the equation using the information of the problem:

Calculating the volume of the total space:

space volume = space long * space wide * space deep

space volume = 84 ft * 42 ft * 9 ft

space volume = 31752 ft3

By converting the volume from ft3 to yd3, we have:

31752 ft3 * (0,037037 yd3 / 1 ft3) = 1176 yd3

Calculating the cost of excavating the volume space:

Total cost = space volume * cost by yard3

Total cost = 1176 yd3 * $39/yd3

Total cost = $45864

<h3>What is volume?</h3>

It is the space occupied by a body, it is calculated by multiplying its dimensions, for example: length, height and width.

Learn more about volume at: brainly.com/question/12628341

#SPJ4

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nata0808 [166]

Answer:

Yes

Explanation:

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3 0
2 years ago
Jan first uses a Michelson interferometer with the 606 nmnm light from a krypton-86 lamp. He displaces the movable mirror away f
gogolik [260]

Answer:

a) d₁ = 247.8 μm

d₂ = 205.3 μm

b) d₂ = 20.53 x 10⁻⁵ m = 205.3 μm

Explanation:

a)

The formula for Michelson Interferometer is derived to be:

d = mλ/2

where,

d = distance moved

m = no. of fringes

λ = wavelength of light

For JAN, we have following data

d = d₁

m = 818

λ = 606 nm = 606 x 10⁻⁹ m

Therefore,

d₁ = (818)(606 x 10⁻⁹ m)/2

<u>d₁ = 24.78 x 10⁻⁵ m = 247.8 μm</u>

For LINDA, we have following data

d = d₂

m = 818

λ = 502 nm = 502 x 10⁻⁹ m

Therefore,

d₂ = (818)(502 x 10⁻⁹ m)/2

<u>d₂ = 20.53 x 10⁻⁵ m = 205.3 μm</u>

b)

The resultant displacement can be found out from the difference between both displacement. And the direction of resultant displacement will be the same as the direction of greater displacement. Therefore,

Resultant Displacement = Δd = d₁ - d₂

Δd = 247.8 μm - 205.3 μm

<u>Δd = 42.5 μm (in the direction of JAN)</u>

4 0
3 years ago
Laboratory experiments here on Earth have determined that each element in the periodic table emits photons only at certain wavel
elena-14-01-66 [18.8K]
The only thing we know about so far that can shift wavelengths of light
to longer wavelengths is when the source of the light is moving away
from the observer.

When we look at the light from distant galaxies, the light from them is
always shifted to longer wavelengths than it SHOULD have. 

AND ... The farther away from us a galaxy IS, the MORE its light is
shifted to wavelengths longer than it should have.

So far, this indicates to us that the whole universe is expanding.
That's the only way to understand what we see, because that's
the only thing we know of that can shift light to longer wavelengths.


By the way ... The most interesting thing about these observations
and measurements is:  When astronomers see this light from distant
galaxies and measure the wavelengths, how do they know how far
the wavelengths shifted ?  How do they know what the wavelengths
SHOULD be ?
 
I'll leave you to read about that in the next few years.
7 0
3 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 1.84 m away from a waterfall 0.251 m in heigh
Liono4ka [1.6K]

Answer:

4.73 m/s

Explanation:

R=1.84  m

H=0.251 m

∅=38.3°

For Horizontal Motion

aₓ=0      uₓ=ucos∅

x=uₓt+\frac{1axt^{2} }{2}

R=ucos∅*t+0

1.84=ucos(38.3)*t

t=\frac{2.344}{u}

For Vertical Motion

Y=Uy*t+\frac{1ayt^{2} }{2}

By putting value

0.251=usin∅(\frac{2.344}{u})-(\frac{1}{2}g *(\frac{2.344}{u} )^{2})

0.251=sin(38.3)(\frac{2.344})-(\frac{1}{2}g *(\frac{2.344}{u} )^{2})

0.251=1.452-\frac{83.057}{u^{2} }

u²=22.41

u=4.73 m/s

7 0
2 years ago
What voltage would be measured across the 15 ohm resistor?
Alona [7]

Answer: That depends on what other components are in the same circuit

along with the 15-ohm resistor.

If there's nothing else, and the battery is connected across the 15-ohm

resistor, then when you measure the voltage across the 15-ohm resistor,

your voltmeter is also connected straight across the battery, and you read

10 volts.

Explanation:

6 0
3 years ago
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