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3 years ago

1. Which of the following is the ideal way to apply pressure onto pedals?

Engineering

2 answers:

Vikki [24]3 years ago

8 0

I think D. By pressing gradually

Galina-37 [17]3 years ago

3 0

Answer:

D. By pressing gradually

Explanation:

When driving, the ideal way to apply pressure onto pedals is by pressing gradually as when you press the pedal more fuel goes into the engine and the speed of the car increases. Because of that, when you are driving it is better to have a steady increase in the pressure because that allows you to maintain the control of the car and avoid risky situations.

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500 flights land each day at San Jose’s airport. Assume that each flight has a 5% chance of being late, independently of whether

BARSIC [14]

Answer:

a.0.0199

b.0.1765

c.0.0785

d.0.1268

e.Yes

Explanation:

It is given that X follows a Binomial distribution with (n= 500, p = 0.05)

The probabilities are computed using the EXCEL .

a) The required probability here is:

P(X less of equal to 15)

= binom.dist(15,500,0.05,TRUE)

=0.0199

Therefore the probability is 0.0199 .

b) The required probability here is:

P(X greater or equal to 30) = 1 - P(X less or equal to 29)

=1 - binom.dist(29,500,0.05,TRUE)

=0.1765

Therefore the probability is 0.1765

c) P(X = 26 )

= binom.dist(26,500,0.05,FALSE)

=0.0785

Therefore the probability is 0.0785

d) The required probability here is computed as:

P(10 less or equal to X less or equal to 20 ) = P(X less or equal to 19) - P(X less or equal to 10)

= binom.dist(19,500,0.05,TRUE) - binom.dist(10,500,0.05,TRUE)

=0.1268

Therefore the probability 0.1268

e) Yes . Therefore the probability because that is the assumption used to apply binomial distribution .

6 0

3 years ago

What is the differences between stack and queue?

Montano1993 [528]

Answer:

A stack is an ordered list of elements where all insertions and deletions are made at the same end, whereas a queue is exactly the opposite of a stack.

Explanation:

7 0

3 years ago

A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in

blondinia [14]

Answer:

(a) The stress on the steel wire is 19,000 Psi

(b) The strain on the steel wire is 0.00063

Explanation:

Given;

length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

extension of the wire, e = 0.75 in

Part (A) The stress on the steel wire;

δ = F/A

= 270 / 0.0144

δ = 18750 lb/in² = 19,000 Psi

Part (B) The strain on the steel wire;

σ = e/ L

L = 100 ft = 1200 in

σ = 0.75 / 1200

σ = 0.00063

Part (C) The modulus of elasticity of the steel

E = δ/σ

= 19,000 / 0.00063

E = 30,000,000 Psi

4 0

3 years ago

Rosita is planning an investigation to determine how a lifeboat's shape affects its

KiRa [710]

Yes! Is right, just did the work!

3 0

3 years ago

Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

Rudik [331]

ANSWERS:

$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

Given:

Piston cylinder assembly which mean that the process is constant pressure process P=C.

AMMONIA

state(1)

saturated vapor $x_{1} =1$

The temperature $T_{1} =0^0 F$

Isothermal process $T=C$

$-V_{2} =2V_{1}$ ( double)

$-V_{2} =.5V_{2}$ (reduced by half)

To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

3 0

3 years ago