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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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In a hydraulic system, accumulator is a device that collects liquid and keeps the liquid under pressure.

Bumek [7]

The answer is: true

6 0

3 years ago

**Please Help. ASAP**

natima [27]

Answer:

The answer is below

Explanation:

1)

$\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at$

2)

$\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)$

3)

$x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}$

4)

$x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}$

5)

$x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}$

6)

$E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}$

7)

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}$

8)

$ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\$

4 0

3 years ago

It is said that Archimedes discovered the buoyancy laws when asked by King Hiero of Syracuse to determine whether his new crown

sertanlavr [38]

Answer with Explanation:

The crown will be pure if it's specific gravity is 19.3

Now by definition of specific gravity it is the ratio between the weight of an object to the weight of water of equal volume

Since it is given that the weight of the crown is 11.8 N we need to find it's volume

Now According to Archimedes principle when the crown is immersed into water the water shall exert a force in upwards direction on the crown with a magnitude equaling to weight of the water displaced by the crown

Mathematically this is the difference between the weight of the crown in air and weight when immersed in water

Thus Buoyant force is $F_{B}=11.8-10.9=0.9N$

Now by Archimedes principle This force equals in magnitude to the weight of water of same volume as of the crown

Thus the specific gravity of the crown equals

$S.G=\frac{11.8}{0.9}=13.11$

As we see that the specific gravity of the crown material is less than that of pure gold hence we conclude that it is impure.

4 0

3 years ago

The constant A in Equation 17.2 is 12π4 R/5(θD) 3 where R is the gas constant and θD is the Debye temperature (K). Estimate θD f

kirill115 [55]

Answer:

The Debye temperature for aluminum is 375.2361 K

Explanation:

Molecular weight of aluminum=26.98 g/mol

T=15 K

The mathematical equation for the specific heat and the absolute temperature is:

$C_{v} =AT^{3}$

Substituting in the expression of the question:

$C_{v} =(\frac{12\pi ^{4}R }{5\theta _{D}^{3} } )T^{3}$

$\theta _{D} =(\frac{12\pi ^{4}RT^{3} }{5C_{v} } ) ^{1/3}$

Here

$C_{v} =4.6\frac{J}{kg-K} *\frac{1kg}{1000g} *\frac{26.98g}{1mol} =0.1241J/mol-K$

Replacing:

$\theta _{D} =(\frac{12\pi ^{4}*8.31*15^{3} }{5*0.1241} )^{1/3} =375.2361K$

3 0

3 years ago

A car moves along a circular track of radius 250 ft such that its seed for a short period of time 0<_ t <_ 4s, is v = 3(t

beks73 [17]

Answer: a) -5 ft/s²

B) 4.5 ft

Explanation: Radius= 250ft

Velocity V = 3(t-t²)ft/s

A). When t= 3 s, the acceleration is

dv/dt = 1-2t

dv/dt = 1- 2(3)

= 1-6

= -5 ft/s²

B. How far it traveled in 3 sec

Distance= 3t²/2 -t³/3 ft

Substituting 3

Distance = 27/2 - 9

= 13.5 - 9

= 4.5 ft

7 0

3 years ago