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3 years ago

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Two very long parallel wires are a distance d apart and carry equal currents in opposite directions. The locations where the net

magnetic field due to these currents is equal to zero are

1 answer:

sergeinik [125]3 years ago

6 0

Answer:

Its not zero anywhere

Explanation:

The magnetic field B at a distance r due to a long conductor carrying current I is given as

B= μol/2pi r

so the net magnetic field due to the current is not zero anywhere

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

6 0

3 years ago

What are the 6 characteristics of life?

stepladder [879]

Organization
Growth and development
Response to a stimulus
Homeostasis
Energy

5 0

3 years ago

Read 2 more answers

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}$

Since

$L_{su}= m_{su}*v_{su}*r_1$

then

$v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}$

$v_{su} = 19.44 m/s$

7 0

4 years ago

A large, simple pendulum is on display in the lobby of the United Nations building. If the pendulum is 18.5 m in length, what is

zhannawk [14.2K]

Answer:

t = 1,144 s

Explanation:

The simple pendulum consists of an inextensible string with a mass at the tip, the angular velocity of this is

w = √( L / g)

The angular velocity is related to the frequency and period

w = 2π f

f = 1 / T

w = 2π / T

Let's replace

2π / T = √ (L / g)

T = 2π √ (g / L)

Let's calculate

T = 2π √ (9.81 / 18.5)

T = 4,576 s

The definition of period in the time it takes the ball to come and go to a given point (a revolution) in our case we go from the end to the middle point that is a quarter of the path

t = T / 4

t = 4,576 / 4

t = 1,144 s

6 0

4 years ago

The universe was 5 percent its current size when light left objects observed now at redshift of ______________

Nuetrik [128]

The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.

The Universe was 5 percent its current size when light left objects now at redshift of <u>19</u>.

Reasons:

The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

$\displaystyle \frac{a}{a_0} =\mathbf{ \frac{1}{1 + z}}$

Where;

a₀ = The current size of the Universe

a = The size of the early Universe = 5% of a

Therefore;

$\displaystyle \frac{a}{a_0} =5\% = 0.05= \frac{1}{1 + z}$

$\displaystyle 0.05 = \frac{1}{1 + z}$

0.05 + 0.05·z = 1

$\displaystyle z = \mathbf{ \frac{1 - 0.05}{0.05} } = 19$

The redshift is of the observed light is, z = <u>19</u>

Learn more here:

brainly.com/question/14459434

brainly.com/question/3654558

4 0

2 years ago