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kompoz [17]
4 years ago
11

A circular loop in the plane of a paper lies in a 0.75 T magnetic field pointing into the paper. The loop's diameter changes fro

m 18.0 cm to 6.8 cm in 0.46 s.
A) Determine the direction of the induced current and justify your answer.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?
Physics
1 answer:
Arte-miy333 [17]4 years ago
4 0

Answer:

Explanation:

A.the direction of induced current will be clockwise

B: Changing 18cm and 6.8cm into 0.18m and 0.68

2.5

Divide them both by 2 to find the radius . Now we have 0.09 and .034m.

Now use Φ=(π*0.09^2)(.75 T)cos0 and the 0.019wb

(π*0.034^2)(.75 T)cos0 and the 0.00272wb

ow use ε=-N(ΔΦ/Δt)

For ΔΦ, 0.091-0.0027=0.0883

C.

To find the current, use I=ε/R

0.0883/2.5= 0.035A

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An airplane is traveling 25° west of north at 300 m/s when a wind with velocity 100 m/s directed 35° east of north begins to blo
hodyreva [135]

Answer:

The resultant velocity is 360.5 m/s  and direction 79° north of east.

Explanation:

Given that,

Velocity of airplane = 300 m/s

Velocity of wind = 100 m/s

Angle θ₁ = 25°

Angle θ₂ =35°

The horizontal velocity component

Using formula of velocity

v_{x}=v_{1}\cos\theta-v_{2}\cos\theta

Put the value into the formula

v_{x}=300\cos65-100\cos55

v_{x}=69.42\ m/s

The vertical velocity component

Using formula of velocity

v_{y}=v_{1}\sin\theta+v_{2}\sin\theta

Put the value into the formula

v_{y}=300\sin65+100\sin55

v_{y}=353.8\ m/s

We need to calculate the resultant velocity

Using formula of resultant velocity

v=\sqrt{v_{x}^2+v_{y}^2}

Put the value into the formula

v=\sqrt{69.42^2+353.8^2}

v=360.5\ m/s

We need to calculate the direction of the resultant velocity

Using formula of direction

\tan\theta=\dfrac{v_{y}}{v_{x}}

Put the value into the formula

\theta=\tan^{-1}(\dfrac{353.8}{69.42})

\theta=79^{\circ}

Hence, The resultant velocity is 360.5 m/s  and direction 79° north of east.

3 0
3 years ago
Name and describe the apparatus used by Cavendish to discover the universal gravitation constant
Evgen [1.6K]
Its like a suspended wood with a lead sphere attached to each of its ends
3 0
4 years ago
What is the minimum value of force acting between two charges placed at 1 m apart from each other?
malfutka [58]

Answer:

Ke²

Explanation:

So,

q1 = e

q2 = e

r = 1m

By coulumb's law,

F = K (q1q2/r²)

F = K (e)(e)/(1)²  

F = Ke²  

 

Option(a)

5 0
3 years ago
A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-ha
Firlakuza [10]

Explanation and answer:

This type of question can be clarified and sometimes solved by drawing a proper diagram or sketch.  (see below)

Solution:

Since we do not know the reaction of the support, we can take moments about the support (thereby eliminating its involvement).

CCW moment = 0.900(5.00/2 - x) kg-m

CW moment = 0.300*(5.00/2-2.00)) = 0.150 kg-m

At equilibrium, CCW moment = CW moment, so

0.900(2.50-x) = 0.150

Expand and solve

2.25 - 0.900x = 0.150

0.900x = 2.25-0.15 = 2.10

x = 2.10 / 0.900 = 2.33 m  (to nearest cm)

8 0
3 years ago
A hungry rat is placed in a maze. It walks the following path to find a piece of cheese. 4.0m N, 7.5 m E, 6.8 m S, 3.7 m E, 3.6
storchak [24]

Answer:

Explanation:

We shall take the help of vector form of displacement . Taking east as i and north as j

4.0m N = 4 j

7.5 m E = 7.5 i

6.8 m S = - 6.8 j

3.7 m E, = 3.7 i

3.6 m S = - 3.6 j

5.3 m W = - 5.3 i

3.7 m N, = 3.7 j

5.6 m W = - 5.6 i

4.4 m S = - 4.4 j

4.9 m W = -  4.9 i

Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i

= -4.6 i -7.1 j

magnitude of displacement = \sqrt{(4.6^2+7.1^2)}

= 8.46 m

Direction

Tanθ = 7.1/ 4.6

θ = 57⁰ south of west .

distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9

= 49.5 m

6 0
3 years ago
Read 2 more answers
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