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3 years ago

A 650 kg steel beam is being pulled up by a crane with a force of 7020 N. What is the upwards acceleration of the beam?

Physics

1 answer:

zimovet [89]3 years ago

8 0

Answer: 1 m/s^2

Explanation:

n= w+ma

n= Mg+Ma

7020 = (650)(9.8)+650a

7020 = 6370+ 650a

650=650a

A= 1 m/s^2

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3 years ago

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konstantin123 [22]

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2 years ago

A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

$A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}$

Differentiate both sides with respect to d.

$\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)$

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

$\frac{d^{2}A}{dt^{2}}= + ve$

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3 years ago

Plz answer brainliest if right!

AleksandrR [38]

Answer: it is D. it is the only possible answer. use the process of elimination. which answers make sense?

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3 years ago

A residential subdivision encompasses 1100 acres with a housing density of four houses per acre. Assume that a high-value reside

aleksandrvk [35]

Answer:

(a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

Explanation:

Given that,

Area = 1100 acres

Number of house in 1 acres = 4

$\text{Number of house in 1100 acres} = 4\times1100$

$\text{Number of house in 1100 acres} = 4400$

Per house water demand = 800 g/day/house

(a). We need to calculate the average daily demand of this subdivision

Using formula for average daily demand

$\text{average daily demand}=house\times\text{Per house water demand}$

$\text{average daily demand}=4400\times800\ gallon/day$

$\text{average daily demand}=3520000\ gallon/day$

$\text{average daily demand}=\dfrac{3520000}{24\times60}\ gallon/min$

$\text{average daily demand}=2444.44\ gallon/min$

The average daily demand of this subdivision is 2444.44 gallon/min.

(b). We need to calculate the design-demand used to design the distribution system

Using formula for the design-demand

$\text{design demand}=(Q_{max})daily\times\text{fire flow}$

$\text{design demand}=1.64\times2444.4\times1000$

$\text{design demand}=4008816\ gallon/m$

Hence, (a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

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3 years ago