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2 years ago

6

In the last 5 meters of braking, you lose ___ of your speed.

1 answer:

expeople1 [14]2 years ago

5 0

Answer:

answer is

a)

3/4

Explanation:

In the last 5 meters of braking, you lose 3/4 of your speed.

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A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2is subjected to an external tensile

lakkis [162]

Answer:

(a) The force sustained by the matrix phase is 1802.35 N

(b) The modulus of elasticity of the composite material in the longitudinal direction Ed is 53.7 GPa

(c) The moduli of elasticity for the fiber and matrix phases is 124.8 GPa and 2.2 GPa respectively

Explanation:

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8 0

3 years ago

The given family of functions is the general solution of the differential equation on the indicated interval.Find a member of th

Alja [10]

Answer:

Explanation:

$y'''+y=0---(i)$

General solution

$y=c_1e^o^x+c_2\cos x +c_3 \sin x\\\\\Rightarrow y=c_1+c_2 \cos x+c_3 \sin x---(ii)\\\\y(\pi)=0\\\\\Rightarrow 0=c_1+c_2\cos (\pi)+c_3\sin (\pi)\\\\\Rightarrow c_1-c_2=0\\\\c_1=c_2---(iii)$

$y'=-c_2\cos x+c_3\cosx\\\\y'(\pi)=2\\\\\Rightarrow2=-c_2\sin(\pi)+c_3\cos(\pi)\\\\\Rightarrow-c_2(0)+c_3(-1)=2\\\\\Rightarrow c_3=-2\\\\y''-c_2\cos x -c_3\sin x\\\\y''(\pi)=-1\\\\\Rightarrow-1=-c_2 \cos (\pi)=c_3\sin(\pi)\\\\\Rightarrow-1=c_2-0\\\\\Rightarrow c_2=-1$

in equation (iii)

$c_1=c_2=-1$

Therefore,

$\large\boxed{y=-1-\cos x-2\sin x}$

5 0

3 years ago

You can change lanes during a turn long as there’s no traffic and you driving slowly

Vanyuwa [196]

Your allowed to switch lanes as long as the road is clear and you use signals.

5 0

2 years ago

At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

ludmilkaskok [199]

Answer:

a) $COP_{R} = 25.014$ , b) $T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

$COP_{R} = 5$

But:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

$T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5} \right)$

$T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

8 0

2 years ago

Read 2 more answers

LINKS GET BODIED ON SITE! RAWR

bulgar [2K]

Answer:

False

Explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZ

7 0

3 years ago