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erica [24]
3 years ago
6

To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.

When the switch is closed, the battery powers two paths in parallel, one of which has a resistor of resistance R1 = 95 Ω in series with an inductor of inductance L = 1.5×10−2 H , while the other has a resistor of resistance R2 = 360 Ω . What is the current supplied by the battery at a time t = 0.4 ms after the switch is closed?

Physics
1 answer:
ki77a [65]3 years ago
3 0

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current i_1. In the case of resistance 2, it will cross a current i_2

Defined this we proceed to obtain our equations,

For i_1,

\frac{di_1}{dt}+i_1R_1 = V

I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

For i_2,

I_2R_2 =V

I_2 = \frac{V}{R_2}

The current in the entire battery is equivalent to,

i_t = I_1+I_2

i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

Our values are,

V=1V

R_1 = 95\Omega

L= 1.5*10^{-2}H

R_2 =360\Omega

Replacing in the current for t= 0.4m/s

i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})

i= 0.0133A

i_1 = 0.01052A

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What affects the amount of potential energy stored in the magnetic field when a magnet is moved against a magnetic force?
LekaFEV [45]

Answer: Strength of magnet and distance from magnetic material

Explanation:

The potential energy of a magnet is determined by the strength of the magnet and the distance between a magnet and another magnet or a magnetic material. Magnetic materials are materials that would be attracted when brought close to a magnet, example of magnetic materials are most metals.

8 0
3 years ago
26. Keenan found the mass of a book to be 4.56*10^ -2 kg . What is the mass of the book in milligrams?​
vagabundo [1.1K]

Taking into account the rule of three for the change of units, the mass of the book is 45600 miligrams.

First of all, the rule of three is a mathematical tool that helps you quickly solve proportionality problems.

Having three known values ​​and one unknown, a proportional relationship is established between all of them in order to find the fourth term of the proportion.

If the relationship between the magnitudes is direct (when one magnitude increases, so does the other; or when one magnitude decreases, so does the other), the rule of three is applied as follows, where a, b and c are known values ​​and x is the unknown to calculate:

a → b

c → x

So: x=\frac{cxb}{a}

Being 1 kg equivalent to 1000000 milligrams, In this case the rule of three is applied as follows: if 1 kg equals 1000000 milligrams, 4.56×10⁻² kg equals how many milligrams?

1 kg → 1000000 milligrams

4.56×10⁻² kg  → x

So:

x=\frac{4.56x10^{-2} kg x1000000 miligrams }{1 kg}

<u><em>x=45600 miligrams</em></u>

In summary, the mass of the book is 45600 miligrams.

Learn more:

  • <u>brainly.com/question/4805238?referrer=searchResults</u>
  • <u>brainly.com/question/5025657?referrer=searchResults</u>
7 0
3 years ago
A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski
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Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0
3 years ago
A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

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