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2 years ago

The density of a certain material is such that it weighs 9 pounds per cubic foot of

Engineering

1 answer:

My name is Ann [436]2 years ago

8 0

Answer:

The answer is "101"

Explanation:

Given value:

We must convert density into pounds per cubic foot to 6900 grams per 4.5 quarter of the length.

$1\ \text{Liquid quart} = 0.03342 \ \text{cubic foot} \\\\1 \ \text{gram} = 0.0022 \ \text{pound}\\$

Formula:

$\bold{density = \frac{6900}{4.5} \frac{gram}{quart}}\\$

$=\frac{6900 \times 0.0022 \ pound }{4.5 \times 0.03342 \cubic \ foot}\\\\= \frac{15.2119}{0.15039} \\\\= 101.1497 \ or \ 101 \ \text{pound per cubic feet}\\$

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Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the o

tresset_1 [31]

Answer:

Vx = 6.242 x 10raised to power 15

Vy = -6.242 x 10raised to power 15

Explanation:

from E = IVt

but V = IR from ohm's law and Q = It from faraday's first law

I = Q/t

E = Q/t x V x t = QV

hence, E =QV

V = E/Q

3 0

2 years ago

Read 2 more answers

49. A solenoid coil with a resistance of 30 ohms and an inductance of 200 milli-henrys, is connected to a 230VAC, 50Hz supply. D

Leya [2.2K]

Answer:

69.59 ohms

Explanation:

Given that

L=200\ mH

F=50\ hZ

R=30\ ohms

For inductance

$X_L=2\times \pi\times f \times LX_L\\=2\times \pi \times 50\times 200\times 10^{-3}\\=62.8\ ohm$

$R=30\ ohmsImpedance\\Z=\sqrt{R^2+X_L^2}\\Z=\sqrt{30^2+62.8^2}\ ohms\\Z =69.59\ ohms$

8 0

2 years ago

1. Add: (i) 5xy, -2xy, -11xy, 8xy (iv) 3a - 2b + c, 5a + 8b -70

Cerrena [4.2K]

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

8 0

1 year ago

An automobile having a mass of 900 kg initially moves along a level highway at 100 km/h relative to the highway. It then climbs

creativ13 [48]

Answer:

ΔKE=-347.278 kJ

ΔPE= 441.45 kJ

Explanation:

given:

mass m=900 kg

the gravitational acceleration g=9.81 m/s^2

the initial velocity $V_{1}$ =100 km/h-->100*10^3/3600=27.78 m/s

height above the highway h=50 m

h1=0m

the final velocity $V_{F}$ =0 m/s

To find:

the change in kinetic energy ΔKE

the change in potential energy ΔPE

assumption:

We take the highway as a datum

solution:

ΔKE=5*m*( $V_{F}$ ^2- $V_{1}$ ^2)

=-347.278 kJ

ΔPE=m*g*(h-h1)

= 441.45 kJ

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2 years ago

A six- lane freeway ( three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hou

Novosadov [1.4K]

Answer:

0.867

Explanation:

The driver population factor ( $f_{p}$ )can be estimated using the equation below:

$f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}$

The value of the heavy vehicle factor ( $f_{HV}$ ) is determined below:

The values of the $E_{T}$ = 2 and $E_{R}$ = 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:

$f_{HV}$ = 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833

Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:

$f_{p}$ = 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867

6 0

2 years ago