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Svetlanka [38]
3 years ago
5

During a 72-ms interval, a change in the current in a primary coil occurs. This change leads to the appearance of a 6.0-mA curre

nt in a nearby secondary coil. The secondary coil is part of a circuit in which the resistance is 12 Ω. The mutual inductance between the two coils is 3.2 mH. What is the change in the primary current?
Physics
1 answer:
Gwar [14]3 years ago
7 0

Answer:

so change in primary current is 1.620 A

Explanation:

Given data

current I =6.0-mA = 6 × 10^(-3) A

resistance R = 12 Ω

mutual inductance  M = 3.2 mH = 3.2 × 10^(-3) H

dt = 72-ms = 72 × 10^-(3) s

to find out

change in the primary current

solution

we know that

Electric and magnetic fields in secondary coil  = mutual inductance × change in primary current / dt      ............1

and we know also that Electric and magnetic fields in secondary coil = resistance × current

so = 6 × 10^(-3) × 12 = 72 × 10^(-3)  volts

so that we say

change in primary current from equation 1

change in primary current  = Electric and magnetic fields in secondary coil × dt / mutual inductance

change in primary current  = 72 × 10^(-3)  ×  72 × 10^(-3) / 3.2 × 10^(-3)

change in primary current  = 1.620

so change in primary current is 1.620 A

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solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum p_{i} before the collision must be equal to the final momentum p_{f} after the collision:

p_{i}=p_{f} (1)

Being:

p_{i}=MV_{i}+mU_{i}

p_{f}=(M+m) V

Where:

M=480 g \frac{1 kg}{1000 g}=0.48 kg the mas of the peregrine falcon

V_{i}=45 m/s the initial speed of the falcon

m=240 g \frac{1 kg}{1000 g}=0.24 kg is the mass of the pigeon

U_{i}=0 m/s the initial speed of the pigeon (at rest)

V the final speed of the system falcon-pigeon

Then:

MV_{i}+mU_{i}=(M+m) V (2)

Finding V:

V=\frac{MV_{i}}{M+m} (3)

V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg} (4)

V=30 m/s (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

F=\frac{\Delta p}{\Delta t} (6)

Where:

F is the force exerted on the pigeon

\Delta t=0.015 s is the time

\Delta p is the pigeon's change in momentum

Then:

\Delta p=p_{f}-p_{i}=mV-mU_{i} (7)

\Delta p=mV (8) Since U_{i}=0

Substituting (8) in (6):

F=\frac{mV}{\Delta t} (9)

F=\frac{(0.24 kg)(30 m/s)}{0.015 s} (10)

Finally:

F=480 N

7 0
3 years ago
PLEASE HELP!!!! WILL GIVE BRAINLIEST!!!!
Svetach [21]
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4 0
3 years ago
Part A.)Six boxes held at rest against identical walls.
krok68 [10]

Answer:

Explanation:

When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .

Hence normal force = reaction force .

From second law also net force is zero , so if  normal force is N and reaction force is R

R - N = mass x acceleration = mass x 0 = 0

R = N .

Ranking normal force from  highest  to smallest

150 N , 130 N , 120 N

B )

Frictional force is equal to the weight of the body because the body is held at rest .

Ranking of frictional force form largest to smallest

7 kg , 5 kg , 3 kg , 1 kg .

Here frictional force is irrespective of the normal force acting on the body  because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .

6 0
3 years ago
A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g
mr Goodwill [35]

The distance an object falls from rest through gravity is

                         D  =  (1/2) (g) (t²)

            Distance  =  (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
if  ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.

                                                      D  =  (1/2) (g) (t²)

Multiply each side by  2 :         2 D  =            g    t² 

Divide each side by ' g ' :      2 D/g =                  t²

Square root each side:        t = √ (2D/g)


Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes: 

-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
                                             and smaller 'g' ==> longer 't' .

-- They don't change by the same factor, because  1/g  is inside
the square root.  So 't' changes the same amount as  √1/g  does.

Gravity on the surface of the moon is roughly  1/6  the value
of gravity on the surface of the Earth.

So we expect ' t ' to increase by  √6  =  2.45 times.

It would take the same bottle  (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
4 0
3 years ago
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Answer:

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