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viva [34]
3 years ago
5

Jamal draws the circuit diagram shown. A rectangular box of lines with the long horizontal side and an extra vertical line at th

e midpoint of the box. The left side has a circle with an X in it and a small circle connected to another small circle by a vertical line. The top side has a small circle connected to another small circle by a horizontal line to the right of the midpoint of the line. The right side has a stack of horizontal lines, which are from top very short, short, very short, short. The bottom line has a zigzag before the midpoint and a zigzag after the midpoint. The extra line has 2 circles with X's in them. How many of each component are shown in the diagram? Select four options. one battery two switches two batteries two resistors three light bulbs three switches
Physics
2 answers:
Natasha2012 [34]3 years ago
7 0

Answer:

one battery

two switches

two resistors

three light bulbs

Explanation: Just got it right :D

vichka [17]3 years ago
4 0

Answer:

BBBBBBBBB

Explanation:

ITS CORRECT

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The rigid beam is supported by the three suspender bars. bars ab and ef are made of aluminum and bar cd is made of steel. if eac
faltersainse [42]

Answer:

Pmax = 67.5 KN

Explanation:

We need to calculate the maximum allowable value of P for both aluminum and steel bars.

<u>FOR STEEL BARS</u>:

Since,

(σallow)st = (Pmax)st/A

where,

(σallow)st = maximum allowable stress of steel bar = 200 MPa = 2 x 10⁸ Pa

A = Cross-sectional area of steel bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)st = Maximum allowable force for steel bar = ?

Therefore,

2 x 10⁸ Pa = (Pmax)st/0.45 x 10⁻³ m²

(Pmax)st = (2 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)st = 9 x 10⁴ N = 90 KN

<u>FOR Aluminum BARS</u>:

Since,

(σallow)al = (Pmax)al/A

where,

(σallow)al = maximum allowable stress of Al bar = 150 MPa = 1.5 x 10⁸ Pa

A = Cross-sectional area of Aluminum bar = 450 mm² = 0.45 x 10⁻³ m²

(Pmax)al = Maximum allowable force for Aluminum bar = ?

Therefore,

1.5 x 10⁸ Pa = (Pmax)al/0.45 x 10⁻³ m²

(Pmax)al = (1.5 x 10⁸ Pa)(0.45 x 10⁻³ m²)

(Pmax)al = 6.75 x 10⁴ N = 67.5 KN

Since,

(Pmax)al < (Pmax)st

Therefore,

The maximum allowable force will be:

Pmax = (Pmax)al

<u>Pmax = 67.5 KN</u>

3 0
4 years ago
PLEASE HELP ME WITH THOSE TWO QUESTIONS
Korolek [52]

Answer:

(a) 0.014 A

(b) 2 V

Explanation:

(a) Applying

V = IR'...................... Equation 1

Where V = Voltge, I = current, R = total resistance.

make I the subject of the equation

I = V/R'................... Equation 2

From the question,

Given: V = 12 V, R' = (122+232+500) ohms (The resistance are connected in series) = 854 ohms

Substitute these values into equation 2

I = 12/854

I = 0.014 A

(b)  Applying

V' = V(R₁)/(R₁+R₂+R₃)...................... Equation 3

Where V' = Voltage across the 100 ohms resistor.

From the question,

V = 12V, R₁ = 100 ohm, R₂ = 200 ohm, R₃ = 300 ohm.

Substitute these values into equation 3

V' = (12×100)/(100+200+300)

V' = 1200/600

V' = 2 V

6 0
3 years ago
Whats the difference between being a christian and being a drug addict..?
tangare [24]
There is a huge diffrenc like one is following christ the other is bringing the devil to pepole
5 0
3 years ago
Read 2 more answers
Que es la energia electrica
Mnenie [13.5K]

Answer:

:) :) :) :) :) :) :) :) :)

Explanation:

:) :) :) .......

6 0
3 years ago
The mean diameters of planets A and B are 8.1 × 103 km and 1.4 × 104 km, respectively. The ratio of the mass of planet A to that
lisabon 2012 [21]

Answer:

dA/dB = 4.955

Approximately, the ratio is 5/1

(Where dA is mean density for planet A while dB is mean density for planet B)

Explanation:

Mass of A = mA

Mass of B = mB

mA/mB = 0.96

Mean radius for A = mA = (8.1 × 10^3)/2 = 4.05 × 10^3 km

Mean radius for B = mB = (1.4 × 10^4)/2

= 7×10^3km

Density = mass/volume

Volume of a sphere = 4/3Πr3

Mean volume for A = (4/3) × Π × (4.05 × 10^3)^3

= 2.784 × 10^11 km3

Mean volume for B = 4/3×Π×(7×10^3)^3

= 1.437 × 10^12km3

Since m/v = d ( where m = mass, v = volume and d = density)

mA = 2.784 × 10^11 km3 × dA ...equation 1

mB= 1.437 × 10^12km3 × dB... equation 2

but mA/mB= 0.96

mA = 0.96 × mB

substitute for mA in equation 1

0.96 × mB = 2.784 × 10^11 x dA equation 3

Substitute for mB in equation 3..

(refer to equation 2)

0.96×1.437×10^12 × dB = 2.784 × 10^11 × dA .....equation 4

divide through by the coefficient of dA

dA = (0.96×1.437×10^12×dB)/(2.784 × 10^11)

divide through by dB

dA/dB = 4.955

therefore, the ratio of dA to dB is 5/1

Therefore, the mean density of A is almost five times that of B

7 0
3 years ago
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