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3 years ago

8

The absence of any mechanical linkage between the throttle pedal and the throttle body requires the use of a _______ motor.

1 answer:

Rom4ik [11]3 years ago

8 0

The answer is A.

hope it helps :)

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A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Put the value into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

$v=ki$

$v=\dfrac{kh}{l}$

Put the value into the formula

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

$V_{s}=\dfrac{M_{s}}{V_{s}}$

Put the value into the formula

$V_{s}=\dfrac{4950\times10^{-3}}{2710}$

$V_{s}=1826.56\ cm^3$

We need to calculate the volume of the soil specimen

Using formula of volume

$V=A\times L$

Put the value into the formula

$V=\dfrac{\pi(17.5)^2}{4}\times17.5$

$V=4209.24\ cm^3$

We need to calculate the volume of the voids

$V_{v}=V-V_{s}$

Put the value into the formula

$V_{v}=4209.24-1826.56$

$V_{v}=2382.68\ cm^3$

We need to calculate the seepage velocity

Using formula of velocity

$Av=A_{v}v_{s}$

$v_{s}=\dfrac{Av}{A_{v}}$

$v_{s}=\dfrac{V}{V_{v}}\times v$

Put the value into the formula

$v_{s}=\dfrac{4209.24}{2382.68}\times0.0187$

$v_{s}=0.0330\ cm/s$

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

8 0

3 years ago

HELP ME PLEASE ASAP!!!!! WILL GIVE BRAINLIEST, FIVE STARS, AND HEART!!!!!<br>(picture included)

harina [27]

Answer:

Kinetic - a box moving

Thermal - sand that feels warm

Electrical - lightning

Radiant - radio waves

Gravitational Potential - fruit hanging from a tree

3 0

2 years ago

Read 2 more answers

Most geologists think that the movement of Earth's plates is cause by

kramer

<span>c) convection currents in the mantle </span>

5 0

3 years ago

Read 2 more answers

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

3 0

3 years ago

A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60

mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

$z(t) = \sqrt{x^{2} + y(t)^{2}}$

$z(t) = \sqrt{70^{2} + 60t^{2}}$

Now, differentiate the above eqn w.r.t 't':

$\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600$

$\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t$

For t = 4 s:

$\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s$

4 0

3 years ago