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3 years ago

Is nickel coin magnetic?

Physics

2 answers:

katen-ka-za [31]3 years ago

8 0

Answer:

Yes

Explanation:

Hey there!

Nickles are always magnetic. So it is the same for a nickle coin.

~I hope I helped you! :)~

Eddi Din [679]3 years ago

3 0

Yes ! The Coin is Magnetic. Try and experiment yourself , make sure a adult is around . Since 1992, coins have been made with steel, which is magnetic. (Pre-1992 coins are made out of bronze, which will not attract.) The dangling shows us how magnetic force travels through some metals, in this case coins (coins become magnetised when they attach) but gets weaker with each one

What the Video for more help : https://terpconnect.umd.edu/~wbreslyn/magnets/is-nickel-magnetic.html

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A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

$L$ = length of the meter stick = 1 m

$m$ = mass of the meter stick

$w$ = angular speed of the meter stick as it hits the floor

$v$ = speed of the other end of the stick

we know that, linear speed and angular speed are related as

$v = r w\\w = \frac{v}{r}$

$h$ = height of center of mass of meter stick above the floor = $\frac{L}{2} = \frac{1}{2} = 0.5 m$

$I$ = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

$I = \frac{mL^{2} }{3}$

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

$(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}$

7 0

3 years ago

Find the ratio of the diameter of iron to copper wire, if they have the same resistance per unit length (as they might in househ

Natasha_Volkova [10]

Answer:

The ratio of the diameter of iron to Cu is;

$\frac{d{Fe} }{ d{Cu} } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}$

Explanation:

R=(ρL)/A

R is resistance,
L is length,
A is area,
ρ is resistivity
d is diameter

from the question the two materials have the same resistance per unit length.

$\frac{R}{L}= \frac{p}{A}$

$\frac{R}{L}$ for iron = $\frac{R}{L}$ for copper

This means we can equate ρ/A for both materials.

$\frac{p_{Fe} }{A_{Fe} } =\frac{p_{Cu} }{A_{Cu} }$

re-arranging the equation we have,

$\frac{A_{Fe}}{A_{Cu} } =\frac{p_{Fe} }{ p_{Cu} }$

$A=\pi \frac{d^{2} }{4}$

$\frac{A_{Fe}}{A_{Cu} } =\frac{d^{2}{Fe} }{ d^{2}{Cu} }$

$\frac{d^{2}{Fe} }{ d^{2}{Cu} } =\frac{p_{Fe} }{ p_{Cu} }$

$\frac{d{Fe} }{ d{Cu} } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}$

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3 years ago

The term water cycle refers to how water moves _____.

allochka39001 [22]

Answer-A

The water cycle shows how water moves around earth.

Ex: Runoffs,precipitation, condensation, evaporation.

-Steel jelly

5 0

3 years ago

Read 2 more answers

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc

Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

$W= F*d= m*g*d$

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

$W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]$

Converting from Joules to Kcals:

$\frac{10620.75}{4186} =2.537 Kcal$

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

x ---> 100%

$x=\frac{2.537*100}{20} =12.685$

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

$\frac{9000Kcals}{12.685Kcals} =709.5 times$

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

$Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\$