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3 years ago

3

When additional resistors are added in parallel to resistors already present in a circuit the equivalent resistance of the the c

ircuit increases. (Section 19.8)
a. True.
b. False.

1 answer:

DiKsa [7]3 years ago

3 0

Answer:

This is the false statement.

Explanation:

In this statement it is said that when additional resistance added in parallel to resistors , the equivalent resistance of the circuit increases which is false statement. Because when an additional resistance is added into the system parallel to the resistors already present inside the circuit the current flow inside the circuit increases. because there are more number of branches formed inside the circuit and due to this current flow inside the circuit increases and the equivalent resistance of the circuit decreases.

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Water flows through a first pipe of diameter 3 inches. If it is desired to use another pipe for the same flow rate such that the

Alborosie

Answer:

the diameter of the second pipe is 2.52 in

Explanation:

Given the data in the question;

We know that; the rate of flow is the same;

so

Av1 = Av2

v ∝ √h

$\frac{A1}{A2}$ = $\frac{V2}{V1}$

$\frac{A1}{A2}$ = √( $\frac{h2}{h1}$ )

( π/4.D1² / π/4.D2² ) = √( $\frac{h2}{h1}$ )

( D1² / D2² ) = √( $\frac{2h1}{h1}$ ) since second is double of first

so

( D1² / D2² ) = √( $\frac{2}{1}$ )

3² / D2² = √2

D2²√2 = 9

D2² = 9/√2

D2² = 9 / 1.4142

D2² = 6.364

D2 = √ 6.364

D2 = 2.52 in

Therefore, the diameter of the second pipe is 2.52 in

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3 years ago

Explain how electric motors convert electrical energy to mechanical energy

alexdok [17]

Answer:

$\rightarrow$ Electric motors build energy through the interaction between magnetic fields.

Magnetic fields may be generated by passing a current through a wire. Winding that wire into a coil intensifies the sector. Wrapping the coil around an iron core can intensify it more. One is going to be of the coil are North, the other South. Reverse the flow of current through the coil. The magnetic poles can swap. Place 2 separate coils close to one another and organize them therefore one turns whereas the opposite is fastened. Then build arrangements for this within the moving coil to reverse even as the opposite poles are about to line up.

Hope it helps!<3

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3 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

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