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velikii [3]
3 years ago
8

Pls help me it’s due today

Engineering
1 answer:
hichkok12 [17]3 years ago
6 0

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

so 14.55

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Burka [1]

Random?

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4 0
3 years ago
Read 2 more answers
I need this asap thank you :) plzzzzz When the spring on a mousetrap car is fully unwound, the force acting on the car is _____.
Usimov [2.4K]

Answer:  (only friction) the friction lets it keep its speed and not slow down and it creats volocity between the serface of where the mousecar is running and the wheels on the ground

sorry if im wrong i tried my best

6 0
3 years ago
While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theo
IgorC [24]

Answer:

V_{c}=1.396x10^{-28}  m^{3} /unit.cell

Explanation:

z = number of atoms

M = Molar mass of zirconium

N = Avogadro’s number

Vc = volume of zirconium unit cell

d = density

z=12x\frac{1}{6}+2x\frac{1}{2}+3=6

z = 6 atoms per unit cell

M = 91.224 g/mol

N = 6.023x10^{23}  atoms/mol

d = 6.51g/cm^{3}

V_{c}=\frac{zxM}{dxN}

V_{c}=\frac{6x(91.224g/mol)}{(6.51g/cm^{3}) x(6.023x10^{23}atoms/mol) }

V_{c}=1.396x10^{-22}  cm^{3} /unit.cell

V_{c}=1.396x10^{-28}  m^{3} /unit.cell

8 0
3 years ago
The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi
Zolol [24]

Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,  

1 gallon = 0.133 cubic feet (cf)

Therefore,  

Capacity of Carvins Cove water reservoir in cf  = 3.2 x 10˄9 x 0.133

                                                                         = 4.28 x 10˄8

 

Applying Mass balance i.e

Accumulation = Mass In - Mass out   (Eq. 01)

Here  

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation  = 0.5 - 11

                         = - 10.5 cfs

 

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

                                                                                       = 37,800

 

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

                                                                                         = 907,200  

 

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days  = (4.28 x 10˄8) x                    1/907,200

                                                                                 = 471 Days.

 

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

8 0
3 years ago
Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Cons
mixas84 [53]

Answer:

work done = 48.88 × 10^{9} J

Explanation:

given data

mass = 100 kN

velocity =  310 m/s

time = 30 min = 1800 s

drag force = 12 kN

descends = 2200 m

to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time  - force ×velocity ×time  - mass ×descends )  10³ J

work done = ( 100 × 310 × 1800  - 12×310 ×1800  - 100 × 2200 )  10³ J

work done = 48.88 × 10^{9} J

6 0
3 years ago
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