All categories

3 years ago

Why were you able to see the effects of the magnetic fields using iron filings

2 answers:

6 0

The iron filings aligned themselves in a pattern, along with the magnet connecting the north and south poles. This creates the “magnetic field”. Interacting with a metal object in between the magnetic properties of the metal interact with the magnet, expressing signs of magnetic force acting on it and the metal, pulled together letting you see the see the effects of the magnetic fields using iron filings.

Talja [164]3 years ago

6 0

Iron filing are soft magnetic materials. The are highly attracted by magnets.
They are used in the study of magnetic fields.
Another reason why they are used because they are light and and they can aline themselves in the paths of the magnetic fields. This way the fields patterns can be seen clearly.

You might be interested in

A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

3 years ago

A ball is thrown down vertically with an initial speed of v0 from a height of h. (a) What is its speed just before it strikes th

Alexus [3.1K]

Answer:

a) v² = v₀² + 2 g h, b) t = v₀/g (1+ √ (1 + 2gh/ v₀²))

Explanation:

a) This is an exercise that we can solve using conservation of energy.

Starting point. High point

Em₀ = K + U = ½ m v₀² + m gh

Final point. Soil

$Em_{f}$ = K = ½ m v²

energy is conserved because there is no friction

Emo = Em_{f}

½ m v₀² + m g h = ½ m v²

v² = v₀² + 2 g h

b) the time it takes to reach the ground can be calculated with kinematics

let's create a reference frame with positive upward direction

v = vo - g t

when it reaches the ground it has a velocity v, the initial velocity is downwards v₀ = -v₀

v = -v₀ - gt

t = - (v + v₀) / g

we substitute the velocity values calculated in the previous part

t = - (√(v₀² + 2 g h) + vo) / g

we will simplify the equation a bit

t = - v₀/g (1+ √ (1 + 2gh/ v₀²))

c) is now thrown vertically upward with the same initial velocity vo.

To find the final velocity we use the conservation of energy where the velocity is squared, so it does not matter if it is positive or negative, therefore in this section the value should be the same as in part a

v = √ (v₀² + 2gh)

d) for this part if there is change since the speed is not squared

v₀ = v₀

v = v₀ - gt

t = (v₀ - v) / g

t = (v₀ - √(v₀² + 2 g h)) / g

t = v₀/g (1 - √(1 + 2gh / v₀²))

3 0

3 years ago

A football player at practice pushes a 60 kg blocking sled across the field at a constant speed. The coefficient of kinetic fric

vichka [17]

Answer:

The force must he apply to the sled is of F= 764.4 N.

Explanation:

m= 60 kg

g= 9.8 m/s²

μ=0.3

W= m*g

W= 588 N

Fr= μ*W

Fr= 176.4 N

F= W + Fr

F= 764.4 N

4 0

3 years ago

Read 2 more answers

How does the end point differ from the equivalence point of a titration?

Gwar [14]

Answer:

Equivalence point and end point are terminologies in pH titrations and they are not the same. 

Explanation:

In a titration the substance added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the acid and base titrated determines the nature of the final solution.

At equivalence point the number of moles of the acid will be equal to the number of moles of the base as given in the equation. The nature of the final solution determines the pH at equivalence point. 

A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating neutral nature of the solution. 

3 0

3 years ago

NEED HELP PLEASEEE 15 POINTS

Nezavi [6.7K]

Answer:

Explanation:

thamjs

7 0

3 years ago

Read 2 more answers