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AlladinOne [14]
3 years ago
15

Who wants to play fortnite my name (JayIsABot13)

Engineering
2 answers:
Blizzard [7]3 years ago
3 0

Answer:

I'll play

Explanation:

I'm insane like so good at it

agasfer [191]3 years ago
3 0

Answer: me ru a default bc I don’t be friends with defaults

Explanation:

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The 150-lb man sits in the center of the boat, which has a uniform width and a weight per linear foot of 3 lb>ft. Determine t
irina1246 [14]

Answer:

M = 281.25 lb*ft

Explanation:

Given

W<em>man</em> = 150 lb

Weight per linear foot of the boat: q = 3 lb/ft

L = 15.00 m

M<em>max</em> = ?

Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):

∑ Fy = 0  (+↑)     ⇒    q'*L - W - q*L = 0

⇒       q' = (W + q*L) / L

⇒       q' = (150 lb + 3 lb/ft*15 ft) / 15 ft

⇒       q' = 13 lb/ft   (+↑)

The free body diagram of the boat is shown in the pic.

Then, we apply the following equation

q(x) = (13 - 3) = 10   (+↑)

V(x) = ∫q(x) dx = ∫10 dx = 10x   (0 ≤ x ≤ 7.5)

M(x) = ∫10x dx = 5x²  (0 ≤ x ≤ 7.5)

The maximum internal bending moment occurs when x = 7.5 ft

then

M(7.5) = 5(7.5)² = 281.25 lb*ft

8 0
4 years ago
The question “How do plants convert sunlight to energy?” best represents which of the following?
OverLord2011 [107]

Answer:

D

Explanation:

I would say this awnser because its the only one that makes sence to me

5 0
3 years ago
A world class runner can run long distances at a pace of 15 km/hour. That runner expends 800 kilocalories of energy per hour. a)
maks197457 [2]

Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars

Explanation :

A)

Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}

= 3780kJ

And 1 hour = 3600s

Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W

B)

At 15km/hour a 15km run takes 1 hour.

1 hour is 3600s and the runner burns 1050 joule per second.

Energy used in 1 hour = 3600 x 1050 J/s

= 3780000 J or 3.78MJ

C)

1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km

15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ

Finally,

1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ

This means that the runner needs 5320/1008 = 5.3 bars

7 0
4 years ago
A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s
monitta

Answer:

a. \dfrac{D_{1}}{ D_{2}}  =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n} which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

v_1 = \dfrac{4}{3} \times  \pi \times r^3

\therefore v_1 = \dfrac{4}{3} \times  \pi \times  \left (\dfrac{10}{2}  \right )^3 = 523.6 \ m^3

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} = \left (\dfrac{T_{1}}{T_{2}}   \right )^{\dfrac{n}{n-1}}

We have;

\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_2}{2}  \right )^3}{\dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_1}{2}  \right )^3}   \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{   \left{D_2}  }{ {D_1}}   \right )^{3\times n} =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n}

\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2  \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

log  \left (\dfrac{D_{1}}{ D_{2}}\right )  =  -3\times n \times log\left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

\left (\dfrac{628.32 }{523.6}   \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{1}{4}}

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}

p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} } =  \dfrac{100\times 10^3}{ \left (1.2) \right  ^{-\dfrac{1}{3} } }

p₂ =  100000/0.941 = 106.265 kPa

W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3  \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

4 0
3 years ago
Explain mathematically whether the divergence of the curl of a vector field is
svp [43]

Answer:

yes

Explanation:

6 0
3 years ago
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