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Juli2301 [7.4K]
3 years ago
8

The ratio of the weight of a substance to the weight of equal volume of water is known as a) Density b) specific gravity c) spec

ific weight d) bulk modulus
Engineering
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

c)specific weight

Explanation:

the ratio between the weight of a substance and the weight of the same volume of water is known as specific weight  this is a property that results in a dimensionless number since it consists of dividing the weight of a fluid by the weight of water in a same volume

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A resonant six-turn loop of closely spaced turns is operating at 50 MHz. The radius of the loop is λ/30, and the loop is connect
Vinvika [58]

Answer:

a) the directivity of the antenna is evaluated to be = 1.761db

b) radiation efficiency is evaluated to be  = 98.67%

c) reflection efficiency gain is evaluated to be =  68%

Explanation:

kindly check the attachment for step to step explanations for better understanding.

6 0
3 years ago
An approach to a signalized intersection has a saturation flow rate of 1800 veh/h. At the beginning of an effective red, there a
Tcecarenko [31]

Answer: The total vehicle delay is

39sec/veh

Explanation: we shall define only the values that are important to this question, so that the solution will be very clear for your understanding.

Effective red time (r) = 25sec

Arrival rate (A) = 900veh/h = 0.25veh/sec

Departure rate (D) = 1800veh/h = 0.5veh/sec

STEP1: FIND THE TRAFFIC INTENSITY (p)

p = A ÷ D

p = 0.25 ÷ 0.5 = 0.5

STEP 2: FIND THE TOTAL VEHICLE DELAY AFTER ONE CYCLE

The total vehicle delay is how long it will take a vehicle to wait on the queue, before passing.

Dt = (A × r^2) ÷ 2(1 - p)

Dt = (0.25 × 25^2) ÷ 2(1 - 0.5)

Dt = 156.25 ÷ 4 = 39.0625

Therefore the total vehicle delay after one cycle is;

Dt = 39

4 0
3 years ago
It is known that the connecting rod AB exerts on the crank BCa 2.5-kN force directed down andto the left along the centerline of
monitta

Answer:

M_c = 61.6 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two casesshown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                         M_c = F_a*( 42 / 150 ) *88

                         M_c = 2.5*( 42 / 150 ) *88

                         M_c = 61.6 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

6 0
3 years ago
Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25
Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx     (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x

y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}

dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx

dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)

dy = \frac{1}{2\pi}

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

#SPJ1

4 0
2 years ago
Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and
Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in  

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that   PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒  P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0
4 years ago
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