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Studentka2010 [4]

3 years ago

11

Car insurance incentives and discounts are available depending on _____. A. school attendance and driver skill B. vehicle type a

nd driver family size C. location and driver employment status D. location, vehicle type, and driving habits

2 answers:

WARRIOR [948]3 years ago

5 0

Answer:D. Location, vehicle type, and driving habits

BaLLatris [955]3 years ago

3 0

D. Location, vehicle type, and driving habits.

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The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E and F. The b

Wewaii [24]

Answer:

a) P ≥ 22.164 Kips

b) Q = 5.4 Kips

Explanation:

GIven

W = 18 Kips

μ₁ = 0.30

μ₂ = 0.60

a) P = ?

We get F₁ and F₂ as follows:

F₁ = μ₁*W = 0.30*18 Kips = 5.4 Kips

F₂ = μ₂*Nef = 0.6*Nef

Then, we apply

∑Fy = 0 (+↑)

Nef*Cos 12º - F₂*Sin 12º = W

⇒ Nef*Cos 12º - (0.6*Nef)*Sin 12º = 18

⇒ Nef = 21.09 Kips

Wedge moves if

P ≥ F₁ + F₂*Cos 12º + Nef*Sin 12º

⇒ P ≥ 5.4 Kips + 0.6*21.09 Kips*Cos 12º + 21.09 Kips*Sin 12º

⇒ P ≥ 22.164 Kips

b) For the static equilibrium of base plate

Q = F₁ = 5.4 Kips

We can see the pic shown in order to understand the question.

7 0

3 years ago

Read 2 more answers

What’s the purpose of current tracks

Taya2010 [7]

Answer:

What's the purpose of tracks going in the red? Having tracks go into the red is surely redundant, I don't see any purpose in having tracks distort ... It just seems like a hang on from the old days of tape, it's something that people who ... be in daws and I'm trying to assemble an alternative I understand the current mixing system. The Dow Jones Industrial Average (DJIA) is a stock index of 30 blue-chip industrial ... Today, the DJIA is a benchmark that tracks American stocks that are ... To calculate the DJIA, the current prices of the 30 stocks that make up the ... the longevity of the Dow serves this purpose better than all other indices.

Explanation:

3 0

3 years ago

A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a slow, constant spee

Ludmilka [50]

Answer:

S = 0.5 km

velocity of motorist = 42.857 km/h

Explanation:

given data

speed = 70 km/h

accelerates uniformly = 90 km/h

time = 8 s

overtakes motorist = 42 s

solution

we know initial velocity u1 of police = 0

final velocity u2 = 90 km/h = 25 mps

we apply here equation of motion

u2 = u1 + at

so acceleration a will be

a = $\frac{25-0}{8}$

a = 3.125 m/s²

so

distance will be

S1 = 0.5 × a × t²

S1 = 100 m = 0.1 km

and

S2 = u2 × t

S2 = 25 × 16

S2 = 400 m = 0.4 km

so total distance travel by police

S = S1 + S2

S = 0.1 + 0.4

S = 0.5 km

and

when motorist travel with uniform velocity

than total time = 42 s

so velocity of motorist will be

velocity of motorist = $\frac{S}{t}$

velocity of motorist = $\frac{500}{42}$

velocity of motorist = 42.857 km/h

3 0

3 years ago

Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr

son4ous [18]

Answer:

The entropy change of the air is $0.240kJ/kgK$

Explanation:

$T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa$

$T_{2}$ is unknown

we can apply the following expression to find $T_{2}$

$-w_{out} =mc_{v} (T_{2} -T_{1} )$

$T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }$

now substitute

$T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K$

To find entropy change of the air we can apply the ideal gas relationship

Δ $s_{air}$ $=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }$

Δ $s_{air} =$ $1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )$

Δ $s_{air} =0.240kJ/kgK$

4 0

3 years ago

In a certain chemical plant, a closed tank contains ethyl alcohol to a depth of 71 ft. Air at a pressure of 17 psi fills the gap

Yuliya22 [10]

Answer:

the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

Explanation:

Given that;

depth 1 = 71 ft

depth 2 = 10 ft

pressure p = 17 psi = 2448 lb/ft²

depth h = 71 ft - 10 ft = 61 ft

we know that;

p = P_air + yh

where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )

so we substitute;

p = 2448 + ( 49.3 × 61 )

= 2448 + 3007.3

= 5455.3 lb/ft³

= 37.88 psi

Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi

5 0

3 years ago