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3 years ago

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31. Three long, straight, parallel wires all lie in the yz plane and each carries a current of 20 A in the positive z direction.

The two outer wires are each 4.0 cm from the center wire. What is the magnitude of the magnetic force on a 50-cm length of either of the outer wires

1 answer:

pochemuha3 years ago

6 0

Answer:

F = 0.5 mN

Explanation:

The full solved solution is in the attach document.

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Whiat is quantum numbers describes the size and energy of an orbital?

Levart [38]

Answer:

The answer is the principal Quantum number (n)

Explanation:

The principal quantum number is one of the four quantum numbers associated with an atom.

It is denoted by a number n=1,2,3,4 etc

It tells both size (directly) and energy (indirectly) of an orbital.

When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.

It tells the energy of the orbital as well as smaller number means less distance from nucleus and having less energy. Since electrons requires to absorb energy to jump into higher orbitals making n=2,3,4 etc. Thus electrons in the orbitals with higher n number indicates higher energy orbitals.

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3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

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3 years ago

A 2.00-kg block of ice is moving on a frictionless horizontal surface. At t = 0 the block is moving to the right with a velocity

enyata [817]

Answer:

Explanation:

a )

We shall apply the concept of impulse .

Impulse = force x time = change in momentum

= 5 x 4 = 2 ( V - 3 ) , where V is final velocity of the object

20 = 2V - 6

V = 13 m /s

b )

Impulse applied = - 7 x 4 = - 28 kg m/s ( negative as direction of force is opposite motion )

If v be the final velocity

2 x 3 - 28 = 2 v ( initial momentum - change in momentum = final momentum )

- 22 = 2v

v = - 11 m /s

object will move with 11 m /s in opposite direction .

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3 years ago

Use the following half-life graph to answer the following question:

Temka [501]

Answer:

A 1.0 min

Explanation:

The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

$m_0 = 50.0 g$

The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

$m'=\frac{50.0 g}{2}=25.0 g$

We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

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3 years ago

When a second student joins the first, the height difference between the liquid levels in the right and left pistons is 40 cm .

vlada-n [284]

Answer:

m = 56.5 kg

Explanation:

Since the addition of mass on one piston caused a change in pressure head at the other. Diameter of the piston calculated is used as 0.46 m

Δm*g / Area = p * g * Δh ..... Eq1

$m = \frac{p*h*A}{g}$

$m = \frac{850 * 0.4 * pi*0.46^2}{4*9.81}\\\\m= 56.5 kg$

8 0

3 years ago