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2 years ago

WILL AWARD BEST ANSWER! find the current flowing across the 30 ohm resistor i = _ a

Physics

1 answer:

Lelu [443]2 years ago

6 0

0.143 A

Explanation:

The total resistance of the circuit is

$r = \frac{(40)(30)}{40 + 30} + \frac{(50)(20 + 10)}{50 + (20 + 10)} \\ = 35.89 \: ohms$

so the total current is

Itotal= V/R = 9V/35.89 ohms

= 0.25 A

To find the voltage drop across the 30-ohns resistor, use

V = Itotal × (combined resistance of 40 30)

= 0.25 A × 17.14 ohms = 4.285 volts

so the current through the 30-ohm resistor is

I = (voltage drop)/30-ohms

= (4.285 V)/(30 ohms)

= 0.143 A

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A ball is thrown horizontally from the top of

Serhud [2]

Answer:

52.38 m/s

Explanation:

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2 years ago

Read 2 more answers

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

2 years ago

Determine the resultant moment of the forces acting about the x, y, and z axes. Suppose that f1 = 200 n, f2 = 60 n, f3 = 115 n,

SpyIntel [72]

Based on the forces acting on the axes, the resultant moments will be (345, 400, 600 N·m)

<h3>What would be resultant moment about x-axis?</h3>

= F₃ x 3

= -115 x 3

= -345 N·m

<h3>What would be resultant moment about y-axis?</h3>

= F₁ x 2

= -200 x 2

= -400 N·m

<h3>What would be the resultant moment about z-axis?</h3>

= F₄ x 2

= -300 x 2

= - 600 N·m

In conclusion, the resultant moment about x, y, and z axes is (345, 400, 600 N·m)

Find out more on resultant moments at brainly.com/question/6278006.

5 0

2 years ago

Two vectors are given by a = 8.6i + 5.1 j and b = 931 + 9.5.

skelet666 [1.2K]

Answer:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>

a) a×b = 34.27k

b) a·b = 128.43

c) (a + b)·b = 305.17

d) The component of a along the direction of b = 9.66

Explanation:

<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:

a) The vectorial product, a×b is:

$a \times b = (8.6*9.5 - 5.1*9.3)k = 34.27k$

b) The escalar product a·b is:

$a\cdot b = (8.6*9.3) + (5.1*9.5) = 128.43$

c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:

$(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17$

d) The component of a along the direction of b is:

$a*cos(\theta) = \frac{a\cdot b}{|b|} = \frac{128.43}{\sqrt{9.3^{2} + 9.5^{2}}} = 9.66$

I hope it helps you!

5 0

3 years ago

Why do heavier objects fall at the same speed as lighter ones?

GaryK [48]

Speed of any freely falling object is always same. Provided, both are left to fall from the same height. If you perform this experiment in a perfect vacuum or near vacuum laboratory, both of them will reach ground with same velocity this is because there is no resistance to their motion. This is always true no matter where you go and perform this experiment.
It can be easily proved from conservation of mechanical energy. Why conserving energy? because there are no forces acting on the freely falling objects other than conservative force(mg).

5 0

3 years ago