Anemia____.<br> help plssssss

develop a plan to test the hardness of a sample of feldspar using the following items glass plate copper penny and streak plate

Artemon [7]

Stroke then 3883 you choke <3333333333

6 0

3 years ago

Name three types of energy that exist in a large piece of charcoal on a grill in the sunlight. Explain why the charcoal has each

zloy xaker [14]

Answer:

* thermal energy

potential energy,

potential energy

Explanation:

The ship has various types of energy,

* thermal energy. This energy is associated with the temperature of the coal, the hotter the greater its internal energy,

* potential energy this energy is stored in the constituent atoms within carbon

* potential energy. It is due to the configuration of the system, in this case the sun heats the coal

5 0

3 years ago

The pink car has a velocity of 4.4 m/s to the right while the blue car has a velocity of 6.8 m/s to the left. If the two cars cr

aleksley [76]

Given :

The pink car has a velocity of 4.4 m/s to the right while the blue car has a velocity of 6.8 m/s to the left.

To Find :

If the two cars crashed and stuck together, what would their combined momentum be afterwards and what direction would they be moving.

Solution :

Let, combined velocity of the car is v.

Also, let us assume that mass of both the cars are equal and it is m.

By conservation of momentum :

Initial momentum = Final momentum

m( 6.8 ) - m( 4.4 ) = 2mv

2mv =2.4m

v = 1.2 m/s

Therefore, the velocity of combined car after crash is 1.2 m/s .

4 0

3 years ago

A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of th

Anon25 [30]

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

8 0

3 years ago

A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1

Pie

Answer:

(a) E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3

(b) $I=0.0085\ A$

Explanation:

Given:

radius if the coil, $r=0.0395\ m$

no. of turns in the coil, $n=520$

variation of the magnetic field in the coil, $B=(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4$

resistor connected to the coil, $R=560\ \Omega$

(a)

we know, according to Faraday's Law:

$emf=n.\frac{d\phi}{dt}$

where:

$d \phi=$ change in associated magnetic flux

$\phi= B.A$

where:

A= area enclosed by the coil

Here

$A=\pi.r^2$

$A=\pi\times 0.0395^2$

$A=0.0049\ m^2$

$\therefore \phi=((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049$

So, emf:

$emf= 520\times \frac{d}{dt} [((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049]$

$emf= 520\times 0.0049\times \frac{d}{dt} [(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)]$

$emf= 2.548\times [0.012+(13.8\times 10^{-5})t^3)]$

$emf= 0.0306+3.516\times 10^{-4}\ t^3$

(b)

Given:

$t_0=5.25\ s$

Now, emf at given time:

$emf=4.7755\times 10^{-2}\ V$

∴Current

$I=\frac{emf}{R}$

$I=\frac{4.7755\times 10^{-2}}{560}$

$I=8.5\times 10^{-5} A$

6 0

4 years ago

Anemia____. help plssssss