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Sidana [21]
3 years ago
9

What can iron filings sprinkled on a piece of paper demonstrate?

Physics
1 answer:
jolli1 [7]3 years ago
6 0
<span>c the pattern of the magnetic fields lines</span>
You might be interested in
Given a 10-V power supply, would a 20-ohm resistor and a 5-ohm resistor need to be arranged in parallel or in series to generate
Bogdan [553]

Answer:

The resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

Explanation:

We are given 10 voltage power source and we have two Resistors with resistance of 20 ohm and 5ohm.

We need to find the orientation in which these two resistors would be arranged so that the circuit could get a current of 2.5Ampere.

Using ohm's law we have

V = I*R

V= voltage

I= current

R= resistance

10 = 2.5*R

R = 10/2.5 = 4ohm

that means we need a total of 4ohm resistance from these two resistors.

since the net Resistance(4ohm) is lower than the smallest resistance(5ohm) available that means the orientation of the resistors will be in parallel.

\frac{1}{R} = \frac{1}{R1} +\frac{1}{R2}\\                 = \frac{1}{20} +\frac{1}{5}\\ \\                 = \frac{1+4}{20} =\frac{1}{4}

R(net) =4ohm

Now the orientation of the resistors are in parallel so the current will be divided.

we know that the current will divide in opposite manner the arm which provides more resistance less current will flow from there and vice versa.

We know that the voltage in parallel remains same

In 20 ohm resistance

again using ohms law

V = i1*R1

10 = i1*20

i1 = 0.5A

in 5ohm resistor

V=i2*R2

10 = I2*5

i2 =2A

and i1+i2 = 0.5+2= 2.5A which means our calculation is correct.

Therefore the resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

6 0
2 years ago
A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k
rewona [7]

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

3 0
3 years ago
What is the modern periodic table of elements?
Kamila [148]
C is the correct answer
3 0
2 years ago
Read 2 more answers
An unknown substance has a mass of 0.125 kg and an initial temperature of 95.0°C. The substance is then dropped into a calorimet
mr Goodwill [35]

Answer:

c = 1163.34 J/kg.°C

Explanation:

Specific heat capacity:

"Specific heat capacity is the amount of heat energy required to raise the temperature of a substance per unit of mass. The specific heat capacity of a material is a physical property."

Use this equation:

mcΔT = ( mw c + mAl cAl )  ΔT'

Rearranging the equation to find the specific heat (c) you get this:

c = (( mw c + mAl cAl )  ΔT') / (mΔT)

c = (( 0.285 (4186) + (0.15)(900)) (32 -25.1)) / ((0.125) (95 - 32))

c = 1163.34 J/kg.°C

7 0
3 years ago
Read 2 more answers
There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.
Sloan [31]

Via half-life equation we have:


A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }


Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes.  By plugging those values in we get:

A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g


There is 6.25 grams left of Ra-229 after 12 minutes.

4 0
3 years ago
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