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horrorfan [7]
3 years ago
6

Find the direction (clockwise or counterclockwise) of the current induced in loop AA if II is steadily decreasing. The current i

n loop AA is clockwise. The current in loop AA is counterclockwise. The current in loop AA is zero.
Physics
1 answer:
IrinaVladis [17]3 years ago
4 0

Answer: counterclockwise

Explanation:

As the area of the loop decreases steadily, the flux through the loop also decreases. By Lenz’s law, any induced current will tend to oppose the decrease. Using Flemming right hand rule (Fleming's right-hand rule which shows the direction of induced current) we know that magnetic field inside the loop due to a counterclockwise current comes out of the plane. Therefore a counterclockwise current will create a stronger magnetic field inside

the loop, tending to increase the flux.

When a wire loop is moved in the direction of the current or a wire loop is being pulled through a uniform magnetic field there would be no induced current ( current loop is said to be zero)

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Katarina [22]

Answer:v=41.23 m/s

Explanation:

Given

mass of heavy object m_1=52 kg

distance of m_1 from the axle r_1=14 cm

mass of rock m_2=123 gm

Length of rod =4.1 m

distance of m_2 from axle r_2=4.1-0.14=3.96 m

Net torque acting is

T_{net}=m_1gr_1-m_2gr_2

T_{net}=52\times 0.14\times g-0.123\times 3.96\times g

T_{net}=6.793\times 9.8

T_{net}=66.57 N-m

Work done by T_{net} is converted to rock kinetic Energy

thus

T_{net}\times \theta =\frac{mv^2}{2}

Where \theta =angle\ turned =\frac{\pi }{2}

v= velocity\ at\ launch

66.57\times \frac{\pi }{2}=\frac{0.123\times v^2}{2}

v^2=66.57\times \pi

v=\sqrt{1700.511}

v=41.23 m/s

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3 years ago
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Answer:

All steps are 20 * 100  (break the rest into appropriate pieces)

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Dominic made the table below to organize his notes about mixtures.
Andrei [34K]

Answer:

Althought it us possible to have more then one state, it's also possible to have only one state of matter

Explanation:

You can make solutions of only one state if matter, for example , it two liquids can be mixed to form a solution they are called miscible.

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mrs_skeptik [129]

Answer:

1. Largest force: C;  smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

F=K\dfrac{ q_{1}q_{2}}{r^{2}}

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write  

F=\dfrac{k}{r^{2}}

1. Net force on each particle

Let's

  • Call the distance between adjacent charges d.
  • Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(b) Force on B

\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}}  - \dfrac{k}{d^{2}}  + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(C) Force on C

\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(d) Force on D

\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

F_{A} : F_{B} : F_{C} : F_{D}  =  \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\

2. Ratio of largest force to smallest

\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}

7 0
3 years ago
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