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evablogger [386]

3 years ago

14

Which of the following statements are true?

1 answer:

olga nikolaevna [1]3 years ago

6 0

C and d are the right answers

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A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I

Basile [38]

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

$E=-\dfrac{d\phi}{dt}$

$E=-B\dfrac{dA}{dt}$

$E=-B\dfrac{A_{2}-A_{1}}{dt}$

$E=B\dfrac{A_{1}-A_{2}}{dt}$ .....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

$l=2\times 2\pi\times r_{1}$

$l=2\times 2\pi\times 0.63$

$l=7.916\ m$

The length when wire is in one loop

$l=2\pi\times r_{2}$

$7.916=2\times \pi\times r_{2}$

$r_{2}=\dfrac{7.916}{2\times \pi}$

$r_{2}=1.259\ m$

We need to calculate the initial area

$A_{1}=N\times\pi\times r_{1}^2$

Put the value into the formula

$A_{1}=2\times3.14\times(0.63)^2$

$A_{1}=2.49\ m^2$

The final area is

$A_{2}=N\times\pi\times r_{2}^2$

$A_{2}=1\times\pi\times(1.259)^2$

$A_{2}=4.98\ m^2$

Put the value of initial area and final area in the equation (I)

$E=0.219\dfrac{2.49-4.98}{0.0572}$

$E=-9.533\ V$

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

6 0

3 years ago

The small particles that produce a streak of light upon entering earth’s atmosphere are called

iris [78.8K]

They are called meteors... i just took the test

8 0

3 years ago

Read 2 more answers

Chris shoots an arrow up into the air. The height of the arrow is given by the function h(t) = - 16t2 + 32t + 26 where t is the

seropon [69]

Answer:

The maximum height of the arrow is 42 (and the units given for the height)

Explanation:

Everything is easier if you make a graph, you can give values to t and replace that values in the function, for example:

When t=0

$h(0)=-16(0^{2})+32(0)+26$

h(0)=26

If you give some values to t you can see how the trajectory of the arrow is (please look the graphic below)

Now, to find the maximum you have to find the derivative of the function that describes the height of the arrow:

$h(t)=-16t^{2}+32t+26$

$h'(t)=-32t+32$

Then you have to take the derivative, and equals to zero to find t:

-32t+32=0

-32t=32

t=1

That is in the time of 1 second the arrow has its maximum height.

Now you have to replace this value in the original function, to find the height of the arrow:

$h(1)=-16(1^{2})+32(1)+26$

h(1)=-16+32+36

h(1)=42

4 0

3 years ago

PLZ HELP DUE TONIGHT

Svetach [21]

By using google, i have found this resource to hep you solve your question.

https://www.siyavula.com/read/science/grade-10/sound/10-sound-03

3 0

3 years ago

6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t

zubka84 [21]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

$d=v\times t$

$d=9.1\times t$ ....(I)

The distance of longitudinal wave

$d=5.7\times (t+71)$ ....(II)

From the first equation

$t=\dfrac{d}{9.1}$

Put the value of t in equation (II)

$d =5.7\times(\dfrac{d}{9.1}+71)$

$\dfrac{9.1d-5.7d}{9.1}=71\times5.7$

$d0.3736=404.7$

$d =1083.24\ km$

Hence, The distance away the center of the earthquake is 1083.24 km.

6 0

3 years ago